text:
\begin{document}

\titlerunning{Viscosity solutions of Hamilton-Jacobi equations in Banach spaces}

\title{On the existence, uniqueness and stability of  $\beta$-viscosity solutions to a class of Hamilton-Jacobi equations in Banach spaces}

\author{Tran Van Bang \and Phan Trong Tien}

\institute{
\begin{itemize}
\item[] 2010 \textit{Mathematics Subject Classification}. 35R15, 49J52, 35D40.
\item[]This research was supported by the Ministry of Education and Training of Vietnam under Grant No. B2018-SP2-14.\\
\item[\Letter] T.V. Bang\\
tranvanbang@hpu2.edu.vn\medskip
\item[] Department of Mathematics, Hanoi Pedagogical University 2, Vinh Phuc, Viet Nam
\item[\Letter] P.T. Tien\\
trongtien2000@gmail.com\medskip
\item[] Department of Mathematics, Quang Binh University, Quang Binh, Viet Nam
\end{itemize}
}

\date{Received: 25 February 2018 / Accepted for publication in AMV: 06 June 2018}

\maketitle

\begin{abstract}
This paper is concerned with the qualitative properties of viscocity solutions
to a class of Hamilton-Jacobi equations (HJEs) in Banach spaces.
Specifically, based on the concept of $\beta$-derivative \cite{DGZ93b}
we establish the existence, uniqueness and stability of  $\beta$-viscosity solutions
for a class of HJEs in the form $u+H(x,u,Du)=0.$
The obtained results in this paper extend ealier works
in the literature, for example, \cite{CL85}, \cite{CL86} and \cite{DGZ93b}.
\keywords{$\beta$-borno, viscosity solutions, Hamilton-Jacobi equations, nonlinear partial differential equations.}
\end{abstract}

\section{Introduction}
Let $X$ be a real Banach space with a $\beta$-smooth norm $|\cdot|$ (see Section 2 for more details), $\Omega\subset X$ an open subset. We study the existence, uniqueness and stability of $\beta$-viscosity solutions for the following HJEs
\begin{equation}\label{A1}
	u+ H(x,u,Du)=0\text{ in }\Omega,
\end{equation}
subject to the boundary condition (in the case $\Omega\not=X$)
 \begin{equation}\label{A2}
 u=\varphi \text{ on }\partial \Omega.
 \end{equation}
Here, $\varphi:\partial\Omega\to \R$ and $H:\overline{\Omega}\times \R\times X^*_\beta\to \R$ are merely continuous in general, where $X^*_\beta$ is the dual space of the Banach space $X,$ and equipped with topology $\tau_{\beta}$ (see Definition 2.2).

Hamilton-Jacobi equations (HJEs) comprise an important class of
nonlinear partial differential equations (PDEs). This type of PDEs
are used to describe dynamics of various practical models and physical processes.
Typical examples of using HJEs in such models can be found
in mechanics or optimal control theory, especially for PDEs.
For instance, in the optimal control theory, a Bellman equation,
also known as a dynamic programming equation, is a necessary condition
for optimality associated with the dynamic programming,
which can be characterized by the so-called Hamilton-Jacobi-Bellman equation,
a subclass of HJEs.  These equations in general, are the nonlinear ones and  often do not have classical solutions, so it is necessary to study weak solutions in some sense.

The concept of viscosity solutions of nonlinear PDEs
was first introduced for HJEs in finite-dimensional spaces
in the earlier of 1980s \cite{dau}. Due to its adaptive capability
in practical applications, particularly in optimal control theory,
this concept has received considerable research attention.
We refer the reader to \cite{AR,H2013} and the references therein.
Recently, the studies on viscosity solutions have also
been developed for HJEs in infinite-dimensional spaces \cite{BZ99,BZ96,CL86,CL85,DGZ93,DGZ93b}
and second-order PDEs both in finite-dimensional spaces \cite{CIL92}
and infinite-dimensional spaces \cite{B2006,BV2006}.
It should be noted that viscosity solutions of PDEs
are weak solutions since they are merely continuous functions whose derivatives
are defined through test functions using extreme principles.
However, it has been shown that some specific viscosity tests
can be conducted through upper and lower differentiations, which are also known as semi-derivatives.
This creates a close connection between the theory of viscosity solutions
and non-smooth analysis including the theory of subdifferentials.

It is known that numerous types of subderivatives (superderivatives)
have been proposed in the literature. Among that the subderivatives
in the sense of  Fr\'{e}chet, Hadamard, G\^{a}teaux and Mordukhovich
are the most widely used ones \cite{BT2014,BZ99,DGZ93b,D2003,MNY2007,MYZ2000}.
Clearly, for a class of HJEs, the use of different subderivatives
may lead to different types of viscosity solutions.
In pioneering works  \cite{AR,CL86,CL85,dau}, the meaning of viscosity solutions
is characterized by semi-Fr\'{e}chet derivative.
On the other hand, in many existing works,
the study on qualitative properties of viscosity solutions is
usally based on a certain type of smoothness of the underlying norm.
However, such smoothness is normally not available in most popular Banach spaces like $L^1.$
To overcome this issue, the authors of \cite{BZ96,DGZ93b}
proposed the concepts of Borno $\beta,$ derivative $\beta$-viscosity, $\beta$-viscosity solutions and obtained a unique result for the HJEs of the form $u+H(x,Du)=0.$

In this paper, we extend the unique result of \cite{BZ96} to a broader class of equations \eqref{A1}, and also  establish the existence and stability of $\beta$-viscosity solutions under certain assumptions. In the literature,
the smooth variational principle proved by Deville in [12] has been employed as a key tool in proving the uniqueness of $\beta$-viscosity solutions of Hamilton-Jacobi equations of the form $u+F(Du)=f$, where $F$ is uniformly continuous on $X^*_\beta$ and $f$ is uniformly continuous and bounded on $X$. However, it is nontrivial to apply that principle for equations of the general form $u+H(x, u, Du)=0$ on a set $\Omega\subset X$. In [8], Crandall and Lions
established the uniqueness of Frechet viscosity solutions for Hamilton-Jacobi equations of the above general form by using the Radon-Nikodym property as a key assumption. By the Radon-Nikodym property, we means that if $\varphi$ is a bounded and lower semicontinuous real-valued function on a closed
ball $B$ in $X$ and $\varepsilon > 0$, then there is an element $x^*\in X^*$
of norm at most $\varepsilon$ such
that $\varphi+x^*$
attains its minimum on $B$. In this work, we use certain extensions of functions to prove the smooth variational principle for functions defined on a subset $E$ of $X$. Then we will establish the uniqueness of $\beta$-viscosity solutions (which are weaker than Fr\'{e}chet-viscosity solutions)   for $u+H(x, u, Du)=0$ on a set $\Omega\subset X$  by the doubling of variables technique. Moreover, we work on a Banach space $X$ having a $\beta$-smooth norm or a norm being equivalent to a $\beta$-smooth norm. The Radon-Nikodym assumption is unnecessary in our arguments. Another contribution of our paper is on the existence and stability of solutions.  More specifically, by Perron's method, we show that under certain conditions, the equations of the form $u+H(x, u, Du)=0$ have $\beta$-viscosity solutions and these solutions as well as the stability of this class of equations.

The remaining of this paper is organized as follows.
Some preliminaries on $\beta$-viscosity subderivatives,
smooth variational principle on Banach spaces and discussion about our main assumptions on the ambient space are presented in Section 2.
Our main results are presented in Section 3. The key result is stated in Theorem \ref{dl3.1} whose proof is postponed to Section 4. From this, the comparison and uniqueness of solutions are proved, as a consequence (Corollary \ref{hq3.1}). The obtained results extend those of  \cite{CL85}, to spaces
endowed with $\beta$-smooth norms, equivalent $\beta$-smooth norms
or the ones satisfying the hypothesis  $(H_{\beta})$ without Radon-Nikodym property. The existence and stability are respectively showed in Theorem \ref{dl3.2} and Theorem \ref{dl3.3}. These are the extension of those in  \cite{CL86} to spaces
endowed with $\beta$-smooth norms instead of Fr\'echet-borno. As mentioned above, Section 4 is used for the proof of Theorem \ref{dl3.1}. In the last section, we give the conclusion and some respects for future study.
\section{Preliminaries}
Let $X$ be a real Banach space with the norm $|.|.$
The close unit ball and the dual space of $X$ will be
denoted as $B_1$ and $X^*$, respectively. We denote by $\left< x^*,x \right>$
the value of  $x^*\in X^*$ at $x\in X;$ that is,  $\left< x^*,x\right> =x^*(x).$ The norm on the dual space $X^*$ will be also denoted as $|.|.$

Throughout this paper, the concerned functions are defined on
the Banach space $X$ taking values in the extended real line $\overline{\mathbb{R}}.$
In addition, it is our convention here that when a function
is lower semicontinuous (resp. upper semicontinuous), it is not identical to $-\infty$ (resp. $+\infty$).

\begin{definition}
A \emph{borno} $\beta$ on $X$ is  a family of closed, bounded, and centrally symmetric subsets of $X$
satisfying the following three conditions:

a) $X=\bigcup\limits_{B\in \beta}B,$

b) $\beta$ is closed under scalar multiplication,

c) the union of any two elements in $\beta$  is contained in some element of $\beta.$
\end{definition}

\begin{example}\label{vd01}
It is easy to verify the following facts.
\begin{enumerate}[1)]
\item The family $F$ of all closed, bounded, and centrally symmetric subsets of $X$ is a borno
on $X,$ which is called \textit{Fr\'{e}chet borno}.
\item The family $H$ of all compact, centrally symmetric subsets of $X$
is a borno on $X$ called \textit{Hadamard borno}.
\item The family $WH$ of all weakly compact, closed, and centrally symmetric subsets of $X$
is a borno on $X$ called \textit{weak Hadamard borno}.
\item The family $G$ of all finite, centrally symmetric subsets of $X$ is also a borno on $X$ called \textit{G\^{a}teaux borno}.
\end{enumerate}
\end{example}

\begin{definition}
A sequence $(f_m)_m\subset X^*$ is said to converge to an $f\in X^*$
in $\beta$-topology $\tau _ {\beta}$ if  $f_m\to f $ when $m\to \infty$
uniformly over all elements of $ \beta,$ this means that
for each set $ M \in \beta $ and any given $\varepsilon>0,$
there exists an $n_0\in \mathbb{N}$ such that $|f_m (x) -f (x)| <\varepsilon$
holds for all $m \ge n_0$ and $x\in M.$
\end{definition}

Suppose that $\beta$ is a  borno  on $X.$ We denote by $\tau_{\beta}$
the topology on $X^*$ associated with the uniform convergence
on $\beta$ subsets and $X^*_{\beta}$ the topological vector space $(X^*, \tau_\beta).$

\begin{remark}\label{nx02}
If  $\beta$ borno  is  $F$  (Fr\'{e}chet), $H$ (Hadamard), $WH$ (Hadamard weak) or $G$ (G\^{a}teaux),
then we have Fr\'{e}chet topology, Hadamard topology, Hadamard weak topology and
G\^{a}teaux topology on the dual space $X^*,$ respectively \cite{DGZ93b}.
Thus, $F$-topology is the strongest topology and $G$ topology is the weakest topology
among $\beta$-topologies on $X^*.$
\end{remark}

\begin{definition}[\cite{BZ96}]
Given a function $f: X \to \overline{\mathbb{R}}.$
We say that $f$ is $\beta$-\textit{differentiable} at $x_0\in X$
with $\beta$-\textit{derivative} $\nabla_\beta f(x_0) = p \in X^*$ if $f(x_0) \in\mathbb{R}$ and
\begin{equation*}\label{eq:01}
\lim_{t\to 0}\frac{f(x_0+th)-f(x_0)-\left<p,th\right>}{t}=0
\end{equation*}
uniformly in $h\in V$ for every $V\in\beta.$

We say that the function $f$ is $\beta$-smooth at $x_0$
if there exists a neighborhood $U$ of $x_0$ such that $f$ is $\beta$-differentiable on $U$
and $\nabla_\beta f: U \to X^*_{\beta}$ is continuous.
\end{definition}

\begin{definition}[\cite{BZ96}]
Let $f: X\to\overline{\R}$ be a lower semicontinuous function and
$f(x)<+\infty.$ We say that $f$ is $\beta$-\textit{viscosity subdifferentiable}
and $x^*$ is a $\beta$-\textit{viscosity subderivative} of $f$ at $x$
if there exists a local Lipschitzian function $g: X\to \mathbb{R}$ such that
$g$ is $\beta$-smooth at $x,$ $ \nabla_\beta g (x) = x^*$ and
$f-g$ attains a local minimum at $ x.$
We denote the set of all $\beta$-subderivatives of $f$ at $x$
by $D^-_{\beta}f(x),$ which is called \textit{$\beta$-viscosity subdifferential} of $f$ at $x.$

Let $f: X\to\overline{\R}$ be an upper semicontinuous function and $f(x)>-\infty.$
We say that $f$ is $\beta$-\textit{viscosity superdifferentiable}
and $x^*$ is a $\beta$-\textit{viscosity superderivative} of $f$ at $x$ if
there exists a local Lipschitzian function $g: X\to \mathbb{R}$ such that
$g$ is $\beta$-smooth at $x,$ $ \nabla_\beta g (x) = x^*$ and $f-g$ attains
a local maximum at $ x.$ We denote the set of all $\beta$-superderivatives of
$f$ at $x$ by $D^+_{\beta}f(x),$ which is called \textit{$\beta$-viscosity superdifferential} of $f$ at $x.$
\end{definition}

Next, we denote
$$\mathcal{D}_{\beta}(X)=\{g: X\to \R| \ g \text{ is bounded, Lipschitzian, and }\beta\text{-differentiable on } X\},$$
$$\|g\|_{\infty}= \sup\{|g(x)|: \ x\in X\},\quad \|\nabla_\beta g\|_{\infty}=\sup\{\|\nabla_\beta g(x)\|: \ x\in X\}$$
and
$$ \mathcal{D}^*_{\beta}(X)=\{g\in \mathcal{D}_{\beta}(X)| \ \nabla_\beta g:X\to X_{\beta}^* \text{ is  continuous}\}.$$

The following hypotheses will be used in the derivation of our results.
\begin{itemize}
\item[$(H_{\beta})$] There exists a  bump function $b$ such that $b\in \mathcal{D}_{\beta}(X);$ and
\item[$(H^*_{\beta})$] There exists a  bump function $b$ (i.e. its support is nonempty and bounded)
such that $b\in \mathcal{D}^*_{\beta}(X).$
\end{itemize}

\begin{proposition}\label{mdkg}
The hypotheses $(H_{\beta}) $ and $(H^*_{\beta})$ are fulfilled
if the Banach space $X$ has a $\beta$-smooth norm.
\end{proposition}

\begin{proof}
Let $f: \mathbb{R} \to \mathbb{R}$ be a continuously differentiable function
with bounded derivative whose support is confined in the interval $[1,3].$
For instance, such a function $f$ is given by
$$f(t)=\begin{cases}\exp\left(\frac{1}{(t-2)^2-1}\right)& \textit{ if }\  |t-2|< 1,\\
0&\textit{ if }\  |t-2|\geq1.
\end{cases}$$
We define the function $b(x)=f(|x|)$ for $x\in X.$
Clearly $b$ is bounded and Lipschitz continuous.
For $|x|\leq 1$ or $|x|\geq3,$ $b(x)=0.$ Due to the $\beta$-smoothness of the norm,
$b$ is differentiable at all points satisfying $1<|x|<3.$
Combining with the smoothness of  $f,$ we can conclude that $(H_{\beta}) $ and $(H^*_{\beta})$ are satisfied.\hfill $\square$
\end{proof}

In general, the converse of
Proposition \ref{mdkg} does not hold. A counterexample can be found in \cite[Page 59]{Ha2},
where the author pointed out that
there exists a Lipschitz continuous bump function, which is Fr\'{e}chet-differentiable
 in a Banach space $X$ without any  G\^{a}teaux-differentiable norm (hence, there does not exist Fr\'{e}chet-differentiable norm).

In \cite{DGZ93}, the authors established a smooth variational principle in a Banach space $X$
where there exist Lipschitz, bounded and differentiable bump functions.
That principle is stated in the following proposition.

\begin{proposition}[Remark II.5, \cite{DGZ93}]\label{dlbp}
Let $X$ be a Banach space satisfying the hypotheses $(H_{\beta})$ (resp. $(H^*_{\beta})$),and
$f$ a lower semicontinuous function on $X.$
Then there  exists a constant $a=a_X>0$
such that for any $\varepsilon \in (0,1)$ and $y_0\in X$ satisfying $f(y_0)<\inf_{X}(f)+a\varepsilon^2,$
there exist a $g\in \mathcal{D}_{\beta}(X)$ (resp $g\in\mathcal{D}^*_{\beta}(X)$) and an $x_0\in X$ such that:
\begin{enumerate}[(a)]
	\item $f+g$ attains its minimum at $x_0.$
	\item $\|g\|_{\infty}\leq \varepsilon$ and $\|\nabla_{\beta}g\|_{\infty}\leq \varepsilon.$
	\item $|x_0-y_0|\le \varepsilon.$
\end{enumerate}
\end{proposition}

From Proposition \ref{dlbp} we have the following result.

\begin{proposition}\label{dlbp1}
Let $X$ be a Banach space satisfying $(H_{\beta})$ (resp. $(H^*_{\beta})$)
and $E$ a closed subset of  $X.$ Then, for a lower semicontinuous bounded from below function $f$
on $E$ and any $\varepsilon \in (0,1),$ there exist
a $g\in \mathcal{D}_{\beta}(X)$ (resp. $g\in\mathcal{D}^*_{\beta}(X)$) and an $x_0\in E$ such that:
\begin{enumerate}[(a)]
	\item $f+g$ attains its minimum at $x_0.$
	\item $\|g\|_{\infty}\leq \varepsilon$ and $\|\nabla_{\beta}g\|_{\infty}\leq \varepsilon.$
\end{enumerate}
\end{proposition}

\begin{proof}
We extend the function $f$ to the function $\widetilde{f}$ defined on $X$ by
$$\widetilde{f}(x)=\begin{cases}	f(x)& \text{ if } x\in E,\\	\infty& \text{ if } x\notin E.\end{cases}$$
Since $E$ is  closed, it follows from the properties of $f$ that
$\widetilde{f}$ is lower semicontinuous and bounded from below.
Thus, for any $\varepsilon>0,$ there exists an $y_0\in X$
such that $\widetilde{f}(y_0)<\inf_X(\widetilde{f})+\varepsilon.$
Moreover, $\widetilde{f}(y_0)<\infty$ and $y_0\in E$ since $\inf_X(\widetilde{f})+\varepsilon<\infty.$
According to Proposition \ref{dlbp}, there exist a $g\in \mathcal{D}_{\beta}(X)$
(resp. $g\in\mathcal{D}^*_{\beta}(X)$) and an $x_0\in X$ such that:
\begin{enumerate}[(a)]
\item $\widetilde{f}+g$  attains its minimum at $x_0.$
\item $\|g\|_{\infty}\le \varepsilon$ and $\|\nabla_{\beta}g\|_{\infty}\le \varepsilon.$
\end{enumerate}

It follows from (a) that $(\widetilde{f}+g)(y)\geq (\widetilde{f}+g)(x_0)$
for any $y\in X.$ In particular, for $y\in E,$ we have $\infty> (\widetilde{f}+g)(y)=(f+g)(y)\geq (\widetilde{f}+g)(x_0).$
Since $\|g\|_{\infty}\leq \varepsilon,$ we can conclude that $x_0\in E.$
The proof is completed.\hfill $\square$
\end{proof}

Similarly, when $f$  is upper semicontinuous and bounded from above on $E,$
the function $\widehat{f}=-f$ is lower semicontinuous and bounded from below on $E.$
We have the
following result.

\begin{proposition}\label{dlbp2}
Let $X$ be a Banach space satisfying $(H_{\beta})$ (resp. $(H^*_{\beta})$)
and $E$ a closed subset of  $X.$ For an upper semicontinuous and bounded from above function $f$ on $E$ and any $\varepsilon \in (0,1),$ there exist a $g\in \mathcal{D}_{\beta}(X)$ (resp. $g\in\mathcal{D}^*_{\beta}(X)$) and an $x_0\in E$ such that:
\begin{enumerate}[(a)]
\item $f-g$ attains its maximum at $x_0.$
\item $\|g\|_{\infty}\le \varepsilon$ and $\|\nabla_{\beta}g\|_{\infty}\le \varepsilon.$
\end{enumerate}
\end{proposition}

\section{The main results}

\begin{definition}
A function $u: \Omega\to\R$ is said to be
\begin{itemize}
\item[(i)] a $\beta$-viscosity subsolution of  \eqref{A1}
if $u$ is upper semicontinuous and for any $x\in \Omega,$ $x^*\in D^+_{\beta} u(x),$ $ u(x)+H(x, u(x), x^*)\le 0;$
\item[(ii)] a $\beta$-viscosity supersolution of (\ref{A1})
if $u$ is lower semicontinuous and  for any $x\in \Omega,$ $x^*\in D^-_{\beta} u(x),$ $u(x)+H(x, u(x), x^*)\geq 0;$
\item[(iii)] a $\beta$-viscosity solution of  (\ref{A1}) if  $u$ is simultaneously
a $\beta$-viscosity subsolution and a $\beta$-viscosity supersolution.
\end{itemize}
\end{definition}

 For convenience, hereafter, we will use the phrases ``$\beta$-viscosity solution of $u+H(x,u,Du)\leq 0$"
and ``$\beta$-viscosity subsolution of $u+H(x,u,Du)=0$" interchangeably. Similarly for the phrases ``$\beta$-viscosity solution of $u+H(x,u,Du)\geq 0$" and ``$\beta$-viscosity supersolution of $u+H(x,u,Du)=0$".

\begin{definition}
A function $u: \Omega\to\R$ is said to be a $\beta$-viscosity subsolution (resp. supersolution, solution) of the problem \eqref{A1}-\eqref{A2} if $u$ is a $\beta$-viscosity subsolution (resp. supersolution, solution) of Equation \eqref{A1} and $u\le \varphi$ (resp. $u\ge \varphi, u=\varphi$) on $\partial \Omega.$
\end{definition}

First, let us recall some auxiliary results. A function $m: [0,\infty) \to  [0,\infty)$
is called a modulus if it is continuous, nonnegative nondecreasing, subadditive on
$[0,\infty)$, and $m(0)=0.$ We will also say that a function
$\sigma: [0,\infty)\times[0,\infty)\to [0,\infty)$ is local modulus if
the function $\sigma(r,R)$ is a modulus for each $R\geq 0$ and $\sigma(r,R)$
is continuous and nondecreasing in both variables.

\begin{remark}\label{nxmd}
It should be noted that if $m: [0,+\infty)\to [0,+\infty)$ is a modulus
then there exist real scalars $A, B$ such that $m(x)\le A+Bx$ for all $x\in [0,+\infty).$
Indeed, for any $x\in [0,+\infty),$ let $[x]$ be the integer part of $x$ and
$A$ is the maximum value of $m$ on $[0,1]$ then $m(x- [x])\leq A.$
Therefore,
\begin{align*}
 m(x)&= m(\underbrace{1+1+\ldots+1}_{ [x] \text{ numbers }}+x- [x])  \\
&\leq [x]m(1)+m(x- [x])\le m(1)x+A=A+Bx,
\end{align*}
where $B=m(1).$
\end{remark}

Next, we make the following assumptions on the function $H.$
\begin{itemize}
\item (H0) There exists a continuous function $w_R: X^*_{\beta} \to \mathbb{R}$ for each $R>0,$ satisfying
\begin{gather*}
|H(x,r,p)-H(x,r,q)|\le w_R(p-q)
\end{gather*}
for any $x\in X,$ $p,q\in X^*$ and $r\in \R$ such that $|x|, |q|, |p|\leq R.$
\item (H1) For each $(x,p)\in X\times X^*,$ $r\mapsto H(x,r,p)$ is nondecreasing.
\item(H1)* For each $(x,p)\in X\times X^*,$ $r\mapsto H(x,r,p)$ is Lipschitz continuous with constant $L_H<1.$
\item (H2) There is a local modulus $\sigma_H$ such that
\begin{gather*}
H(x,r,p)-H(x,r,p+q)\le \sigma_H(|q|,|p|+|q|)
  \end{gather*}
for all $r\in \mathbb{R},$ $x\in\Omega $ and $p, q\in X^*.$
\item (H3) There is a modulus $m_H$ such that
 \begin{equation}
 \begin{split}
H(y,r,\lambda (\nabla_{\beta}|.|^2)(x-y))-&H(x,r,\lambda (\nabla_{\beta}|.|^2)(x-y))\\
&\leq m_H(\lambda |x-y|^2+ |x-y|)
 \end{split}  \notag
\end{equation}
for all $x,y\in \Omega$ with $x\neq y,$
$r\in\R$ and $\lambda\geq 0.$
\end{itemize}

\begin{example}
Suppose that $X$ is a Banach space with a $\beta$-smooth norm.
Let $b: X\to X$ be a bounded and Lipschitz continuous function. Consider the following functions:
$$H(x,r,p)=\langle p,b(x)\rangle \text{ and } H^*(x,r,p)=-\frac{1}{2}r+\langle p,b(x)\rangle.$$
It can be verified that the function $H$ satisfies hypotheses (H0), (H1), (H2) and (H3). Specifically, if $b$ is bounded on $X$ by a constant $K,$ and its Lipschitz constant is $L_b,$ then
\begin{align*}
|H(x,r,p)-H(x,r,q)|&=|\langle p-q,b(x)\rangle| \le |p-q|.|b(x)|\\
&\le K|p-q|=: w_R(p-q)
\end{align*}
for any $x\in X,$ $p,q\in X^*,r\in \R$ satisfying $|x|, |q|, |p|\le R.$
Thus, (H0) is satisfied. Moreover, (H1) and (H2) are obvious. We also have
 \begin{align*}
  H(y,r,\lambda (\nabla_{\beta}|.|^2)(x-y))&-H(x,r,\lambda (\nabla_{\beta}|.|^2)(x-y))\\
 &=\langle\lambda (\nabla_{\beta}|.|^2)(x-y),b(y)-b(x)\rangle\\
&\leq 4KL_b\lambda |x-y|^2\\
&\leq 4KL_b(\lambda |x-y|^2+|x-y|)=: m_H(\lambda |x-y|^2+ |x-y|).
 \end{align*}
Therefore, (H3) is fulfilled.

Similarly, the function $H^*$ satisfies the hypotheses (H0), (H2), (H3). It should be emphasized that the function $H^*$ does not satisfy  (H1) while it satisfies $(H1)^*$.
\end{example}

On the basis of the above preliminaries, we are now in a position to present our main result on the uniqueness of $\beta$-viscosity solutions
of \eqref{A1}.

\begin{theorem}\label{dl3.1}
Let $X$ be a Banach space with a $\beta$-smooth norm, and $\Omega$ a open subset of $X.$
Assume that the function $H$ satisfies assumptions (H0), (H1) (resp. (H1)*), (H2), (H3), and
$\widehat{H}$ satisfies (H0). Let $u,v\in C(\overline{\Omega})$ respectively be
$\beta$-viscosity solutions of the problems
\begin{equation}
u+H(x,u,Du)\leq 0  \text{ and } v+\widehat{H}(x,v,Dv)\ge 0 \text{  on }\Omega,
\label{1.9}
\end{equation}
and suppose that there is a modulus $m$ such that
\begin{equation}\label{1.10}
 |u(x)-u(y)|+ |v(x)-v(y)|\le m(|x-y|)\text{ in } \Omega.
\end{equation}
Then we have
\begin{equation}\label{1.11}
\begin{split}
  u(x)-v(x)&\leq \sup_{\partial \Omega} (u-v)^+
  +\sup_{\Omega\times \R\times X^*}(\widehat{H}-H)^+,\\
 \Big(resp.\quad  u(x)-v(x)&\leq \sup_{\partial \Omega} (u-v)^+
  +\frac{1}{1-L_H}\sup_{\Omega\times \R\times X^*}(\widehat{H}-H)^+\Big),
\end{split}
\end{equation}
for all $x\in \Omega.$

In particular, when $\Omega = X,$ we have the estimate \eqref{1.11} in which the term $\sup_{\partial \Omega} (u-v)^+$ on the right hand side is replaced by zero.
\end{theorem}

The proof of this theorem is rather long. To
facilitate the reading, it is postponed to Section 4.

\begin{corollary}[Comparison and uniqueness]\label{hq3.1}
Given $X$ a Banach space, and equipped with a $\beta$-smooth norm. Let $\Omega\subset X$ be an open set with boundary $\partial\Omega\not=\emptyset,$ $\varphi$ a continuous function on $\partial \Omega.$ Assume that the function $H$ satisfies the assumptions (H0), (H1) (resp. (H1)*), (H2), and (H3). If $u,v\in C(\overline{\Omega})$ respectively are
$\beta$-viscosity subsolution and $\beta$-viscosity supersolution of Equation \eqref{A1} satisfying \eqref{1.10}, then $u\leq v$ in $\Omega,$ provided that $u\leq v$ on $\partial \Omega.$ Therefore, the problem \eqref{A1}, \eqref{A2} has at most a solution in $C(\overline{\Omega}).$

In the case $\Omega$ is the whole space $X,$ the comparison and the uniqueness of the solution for Equation \eqref{A1} is an obvious consequence.
 \end{corollary}
\begin{proof}
The comparison follows immediately from Theorem \ref{dl3.1}. Let $u,v\in C(\overline{\Omega})$ be the solutions of the problem \eqref{A1}, \eqref{A2}. Then, $u$ is a subsolution, $v$ is a supersolution of \eqref{A1} and $u\leq v$ on $\partial \Omega,$ so by the comparison result, $u\leq v$ in $\Omega.$ By changing the role of $u$ and $v,$ we obtain that $u\geq v,$ and therefore $u=v$ in $\Omega.$ \hfill $\square$
\end{proof}

\begin{proposition}\label{md3.1}
If in Corollary \ref{hq3.1}, the space $X$ satisfies either $(H_{\beta})$ or $(H^*_{\beta})$ but it has neither $\beta$-smooth norm  nor  norm being equivalent to a $\beta$-smooth norm, then the conclusions of  Corollary \ref{hq3.1} are not true.
\end{proposition}
\begin{proof}
Indeed, let $H(x,u,Du)=1+u$ in Equation \eqref{A1}. It is easy to  see that the function $u\equiv -\dfrac{1}{2}$ is a $\beta$-viscosity solution of Equation \eqref{A1}. We will show that $u=|x|$ is another solution.
Indeed,  we take the space $X$ satisfying $(H_{\beta})$ and having a norm which is not $\beta$-differentiable at any point (such a space $X$ can be the one that Remark II.9, \cite{DGZ93} has pointed out). Then $u$ is obviously  a $\beta$-viscosity supersolution of Equation \eqref{A1}. On the other hand, since $u=|x|$  is a convex function and not $\beta$-differentiable at any point on  $X,$  $D^+_{\beta}u(x)=\emptyset$ for all $x\in X.$ Thus $u$ is a  $\beta$-viscosity subsolution of equation \eqref{A1}.
\hfill $\square$
\end{proof}

\begin{remark}\label{can1}

	\begin{enumerate}[1)]
		\item When  $\beta =F$ is Fr\'{e}chet borno, we get the result of Crandall M. G. and Lions P. L. in \cite{CL85}.

		\item If  the space $X$ has no $\beta$-smooth norm  but it has a  norm which is equivalent to a $\beta$-smooth norm, we also have the conclusion of  Theorem \ref{dl3.1} by using the equivalent norm instead of the original one.

		\item In \cite[Theorem 3.6]{leduc}, the authors confirmed that the hypothesis of existing a Fr\'{e}chet-smooth bump function   is sufficient for the existence of an equivalent Fr\'{e}chet-smooth norm on space $X.$ Thus, Theorem \ref{dl3.1}  is a generalization of   Crandall's result in \cite{dau}. Note that in this work, we do not use the  Radon-Nikodym property of the space $X.$
		\item By \cite[page 211]{DGZ93}, the assumptions that there exists a  function $d: X\to \mathbb{R}$ which is bounded and Fréchet differentiable on $X\backslash \{0\},$ and there is a real number $r>0$ such that $d(x)\ge r |x|$ used in \cite{dau} are equivalent to the existence of a Lipschitz and Fr\'{e}chet differentiable bump function on $X.$ Hence, if there is a such function $d,$ then Theorem \ref{dl3.1} still holds without assumption on the existence of a Fr\'{e}chet-smooth norm or the Radon-Nikodym property as in \cite{dau}.
	\end{enumerate}
\end{remark}
\begin{proposition}\label{md3.2}
Let $X$ be a normed space  without a norm being equivalent to a $\beta$-smooth norm but satisfies the assumption $(H_{\beta}).$ Then by adding the boundedness of the involving solutions, Theorem \ref{dl3.1} is still valid.
\end{proposition}
\begin{proof}
  By \cite[Lemma 2.15]{BZ96},  there exist a function $d: X\to \mathbb{R^+}$  and  $K>1$ such that
\begin{enumerate}[i)]
	\item $d$ is bounded, Lipschitz continuous on $X$, and $\beta$-smooth on $X\setminus \{0\};$
	\item $|x|\le d(x)\le K |x|$ if $|x|\le 1$ and $d(x)=2$ if $|x|\ge 1.$
\end{enumerate}

Now assume that $u,v$ are bounded on $\Omega.$ In the proof of Theorem \ref{dl3.1}, we replace $|.|^2$ by $d^2(\cdot)$  and consider the function
\begin{equation*}
\Phi(x,y)=u(x)-v(y)-\left(\frac{d(x-y)^2}{\varepsilon}+\lambda\zeta(|x|-R)\right)
\label{1.161}
\end{equation*}
The conclusion follows by using a similar argument as the proof of Theorem \ref{dl3.1}.
\hfill $\square$
\end{proof}

We emphasize that the hypothesis on the boundedness of  $u,v$ in Proposition \ref{md3.2} is necessary. A  counterexample can be found in the proof of Proposition \ref{md3.1}.

 We proceed to study the stability of the $\beta$-viscosity solutions. Using this stability in the same way as in \cite{DGZ93}, we obtain Proposition \ref{dlhb}.

\begin{theorem}[Stability]\label{dl3.2}
	Let $X$ be a Banach space with a $\beta$-smooth norm, and $\Omega$ an  open subset of $X.$
	Let $u_n\in C(\Omega)$ and $H_n\in C(\Omega\times \mathbb{R}\times X^*_{\beta}),$ $n=1, 2, ...$ converge to $u, H$ respectively as $n\to \infty$ in the following way:

	For every $x\in \Omega$ there is an $R>0$ such that $u_n\to u$ uniformly on $B_R(x)$ as $n\to\infty,$  and if $(x,r,p), (x_n,r_n,p_n)\in \Omega\times \mathbb{R}\times X^*_{\beta}$ for $n=1, 2, ...$ and $(x_n,r_n,p_n) \to (x,r,p)$ as $n\to\infty,$ then $H_n(x_n,r_n,p_n)\to H(x,r,p).$
	If $u_n$ is a $\beta$-viscosity supersolution (respectively, subsolution) of $H_n=0$ in $\Omega,$ then $u$
	is a $\beta$-viscosity supersolution (respectively, subsolution) of $H=0$ in $\Omega.$
\end{theorem}
\begin{proof}
	We treat the case that the $u_n$ are $\beta$-viscosity supersolutions of $H_n= 0;$ the case of $\beta$-viscosity subsolutions is entirely similar. Let $x\in \Omega$ and $p\in D^-_{\beta} u(x),$ we can find a local Lipschitz continuous and $\beta$-differentiable function $\varphi: X\to \mathbb{R}$  on $D(\varphi)$ such that:
	\begin{itemize}
		\item[(i)] $(u-\varphi)(x)=0$ and $(u-\varphi)(y)\ge 0$ if $y\in \Omega.$
		\item[(ii)] $p=\nabla_{\beta}\varphi(x)$ and $\nabla_{\beta}\varphi$ is norm to $\tau_{\beta}$ continuous at $x.$
	\end{itemize}
	Since $u_n$ converges to $u$ as $n\to \infty$, then we can assume that there exist a sequence of positive numbers $\{\varepsilon_n\}_n$  such that $\varepsilon_n \to 0$
	as $n\to \infty$ and
	$$u(y)\le u_n(y)+\varepsilon_n \text{ for }y\in \Omega.$$
	Therefore,  $$u_n(y)+\varepsilon_n-\varphi(y)\ge(u-\varphi)(y)\ge 0\text{ for }y\in \Omega,$$
	which implies $\inf (u_n-\varphi)+\varepsilon_n\ge 0.$
	Since $u$ and $\varphi$ are continuous at $x,$ there exist $z_n\in \Omega,$ $|z_n-x|<\varepsilon_n$ such that
	$u_n(z_n)-\varphi(z_n)\le \varepsilon_n$. Thus,  $u_n(z_n)-\varphi(z_n)\le \inf (u_n-\varphi)+2 \varepsilon_n.$ Let us define $(u_n-\varphi)(y)=-\varepsilon_n$ for $y\notin \Omega$. Then $u_n-\varphi$ is lower semicontinuous and bounded from  below. According to Proposition \ref{dlbp}, there exists $\psi_n\in \mathcal{D}_{\beta}(X)$ such that
	\begin{itemize}
		\item[(1)] $\lim\limits_{n\to \infty}||\psi_n||=0$ and $\lim\limits_{n\to \infty}||\nabla_{\beta}\psi_n||_{\infty}=0,$
		\item[(2)] $u_n-\varphi+\psi_n$ attains a local minimum at $y_n\in \Omega,$
		\item[(3)] $\lim\limits_{n\to \infty}|y_n-z_n|=0.$
	\end{itemize}
	Using (3) and the fact that $|z_n-x|<\varepsilon_n,$ we have $\lim\limits_{n\to \infty}|y_n-x|=0.$ On the other hand, $p_n=\nabla_{\beta}\varphi(y_n)-\nabla_{\beta}\psi_n(y_n)\in D^-_{\beta} u_n(y_n) $ and $p_n-p=\nabla_{\beta}\varphi(y_n)-\nabla_{\beta}\varphi(x) -\nabla_{\beta}\psi_n(y_n).$
	Since $\nabla_{\beta}\varphi$ is a continuous function, $y_n\to x$ and $\lim\limits_{n\to \infty}||\nabla_{\beta}\psi_n||_{\infty}=0,$ $p_n-p\to 0$ when $n\to \infty.$

	Now using the fact that $u_n$ is a $\beta$-viscosity supersolution of $H_n=0,$ we obtain
	$$H_n(y_n,u_n(y_n),p_n)\ge 0.$$
	Passing to the limit as $n\to \infty$, we get $H(x,u(x),p)\ge 0.$ Thus, $u$ is $\beta$-viscosity supersolution of $H(x,r,p)=0.$
	\hfill $\square$
\end{proof}

Finally, we close this section with a finding on the existence of $\beta$-viscosity solutions. The following Theorem \ref{dl3.3} generalizes the ones in \cite{CL86} and \cite{DGZ93}. To prepare for that theorem, we recall several definitions and a proposition proved in \cite{DGZ93}.

If $\Omega$ is an open subset of a Banach space $X$ and if $u$ is a function defined on $\Omega,$   the upper semicontinuous envelope $u^*$ of $u$ is defined as
$$u^*=\inf\{v: v\text{ is continuous on }\Omega\text{ and }v\ge u\text{ on }\Omega\}$$
and the lower semicontinuous envelope $u_*$ of $u$ is defined by
$$u_*=\sup\{v: v\text{ is continuous on }\Omega\text{ and }v\le u\text{ on }\Omega\}.$$

\begin{proposition}[Proposition III.7, \cite{DGZ93}]\label{dlhb}
	Let $X$ be a Banach space satisfying $(H^*_{\beta}),$  $\Omega$ an open subset of $X,$  and $F: \overline{\Omega} \times\mathbb{R} \times X^*_\beta \to \mathbb{R}$ continuous. Suppose that $u_0, v_0$ are  respectively a $\beta$-viscosity subsolution and a $\beta$-viscosity supersolution of $F(x,u,Du)=0$ on $\Omega.$ Assume $u_0\le v_0$ on $\Omega.$
	Then, there is a function $u: X\to \mathbb{R}$ such that $u_0\le u\le v_0$ on $\Omega,$ $u^*$ is a $\beta$-viscosity subsolution of $F(x,u,Du)=0$ on $\Omega,$ and $u_*$ is a $\beta$-viscosity supersolution of $F(x,u,Du)=0$ on $\Omega.$
\end{proposition}

We are now ready to establish the existence.

\begin{theorem}[Existence]\label{dl3.3}
	Let $X$ be a Banach space with a $\beta$-smooth norm, and $\Omega$ an  open subset of $X.$ Let $H: \Omega \times\mathbb{R} \times X^* \to \mathbb{R}$ satisfy (H0), (H1) (resp. (H1)*), (H2), (H3), $\varphi\in C(\partial \Omega)$ such that $|\varphi (x)|\le M+N|x|$ for some real numbers $M, N\in (0,\infty)$ and for all $x\in \partial\Omega.$ Then there exists a unique $\beta$-viscosity solution of the problem \eqref{A1}, \eqref{A2}.
\end{theorem}
\begin{proof}
	The uniqueness part of the problem is asserted in	Corollary \ref{hq3.1}. Here we only present the proof of the existence using the Perron's method.

	By taking $\lambda=0$ and $r=0,$ the assumption (H3) gives us $H(x,0,0)- H(0,0,0)\le m_H(|x|),$ so
	$$ H(x,0,0)\le H(0,0,0)+m_H(|x|) \le |H(0,0,0)|+m_H(|x|). $$
Similarly, we have $- H(x,0,0)\le |H(0,0,0)|+m_H(|x|).$ Therefore $|H(x,0,0)|\le |H(0,0,0)|+m_H(|x|).$

	Since $m_H$ is a modulus, according to Remark \ref{nxmd}, there exist real numbers $A_H, B_H$ such that
	$$	|H(x,0,0)|\le A_H+B_H |x|,\quad \forall x\in \Omega.$$

	Let $B_1= \max\{B_H,N\}.$  Defining  $$A_1=\max\{A_H+\sigma_H(B_H,B_H), M\},\quad  A_2=\max\{A_H+\sigma_H(B_1,2B_1), M\},$$
	$$A'_1=\max\Big\{A_H+\sigma_H\Big(\frac{B_1}{1-L_H},\frac{B_1}{1-L_H}\Big), M\Big\};$$ $$A'_2=\max\Big\{A_H+\sigma_H\Big(\frac{B_1}{1-L_H},\frac{2B_1}{1-L_H}\Big), M\Big\}$$
	and
	$$v_0(x)=A_1+B_1|x|,\quad u_0(x)=-(A_2+B_1|x|).$$
	$$\Big({\rm resp.}\quad v_0(x)=\frac{A'_1+B_1|x|}{1-L_H},\quad u_0(x)=-\frac{A'_2+B_1|x|}{1-L_H}\Big).$$

	We claim that $v_0$ is a $\beta$-viscosity supersolution and $u_0$ is a $\beta$-viscosity subsolution of the problem \eqref{A1}, \eqref{A2} in $\Omega.$
	Indeed, it is clear that $v_0\geq \varphi$ on $\partial \Omega.$ Moreover, from the assumptions (H1) (resp. (H1)*) and (H2), for each $x\in \Omega,$ $p\in D^-_{\beta} (|\cdot|)(x),$ we have
	\begin{align*}
	H(x,0,0)&\le H(x,A_1+B_1|x|,0)\\
	& \le H(x,A_1+B_1|x|,B_1p)+\sigma_H(B_1,B_1);
	\end{align*}
	\begin{align*}\Big({\rm resp.} \quad
	H(x,0,0)&\le H(x,v_0(x),0)+L_Hv_0(x)\\
	& \le H\Big(x,v_0(x),\frac{B_1}{1-L_H}p\Big)+\sigma_H\Big(\frac{B_1}{1-L_H},\frac{B_1}{1-L_H}\Big)+L_Hv_0(x)
	 \Big). \end{align*}
	Hence
	\[\begin{split}
	v_0(x)+H(x, v_0(x),B_1p)&\ge A_1+B_1|x|+ H(x,0,0)-\sigma_H(B_1,B_1)\\
	&\ge A_1+B_1|x|- A_H-B_H|x|-\sigma_H(B_1,B_1)\\
	&\ge 0;
	\end{split}\]
	\[\begin{split}
	\Big({\rm resp.}\quad  v_0(x)+&H\Big(x, v_0(x),\frac{B_1}{1-L_H}p\Big)\\
	&\ge (1-L_H)v_0(x)+ H(x,0,0)-\sigma_H\Big(\frac{B_1}{1-L_H},\frac{B_1}{1-L_H}\Big)\\
	&\ge A'_1+B_1|x|- A_H-B_H|x|-\sigma_H\Big(\frac{B_1}{1-L_H},\frac{B_1}{1-L_H}\Big)\ge 0\Big).
	\end{split}\]
	This proves that $v_0$ is a $\beta$-viscosity supersolution of \eqref{A1}.

	On the other hand, by the definition, $u_0\leq\varphi$ on $\partial\Omega.$ It remains to prove that $u_0$ is a $\beta$-viscosity subsolution of \eqref{A1} in $\Omega.$

	From the conditions (H1) (resp. (H1)*) and (H2), for each $x\in \Omega,$ $p\in D^+_{\beta}(|~\cdot~|)(x),$ we have
	\begin{gather*}
	H(x,u_0(x),-B_1p)\le H(x,0,-B_1p)\le H(x,0,0)+\sigma_H(B_1,2B_1).
	\end{gather*}
	\[\begin{split}\Big({\rm resp.}\quad
	H\Big(x,u_0(x),&-\frac{B_1}{1-L_H}p\Big)\le H\Big(x,0,-\frac{B_1}{1-L_H}p\Big)+L_H|u_0(x)|\\
	&\le H(x,0,0)+\sigma_H\Big(\frac{B_1}{1-L_H},\frac{2B_1}{1-L_H}\Big)+L_H|u_0(x)|\Big).
	\end{split}\]
	Hence
	\[\begin{split}
	u_0(x)+H(x,u_0(x),-B_1p)&\le u_0(x)+H(x,0,0)+\sigma_H(B_1,2B_1)\\
	&\le u_0(x)+A_H+B_H|x|+\sigma_H(B_1,2B_1)\le 0;
	\end{split}\]
	\[\begin{split} 	\Big({\rm resp.}\quad
	u_0(x)+&H\Big(x,u_0(x),-\frac{B_1}{1-L_H}p\Big)\\
	&\le (1-L_H)u_0(x)+H(x,0,0)+\sigma_H\Big(\frac{B_1}{1-L_H},\frac{2B_1}{1-L_H}\Big)\\
	&= -(A'_2+B_1|x|)+A_H+B_H|x|+\sigma_H\Big(\frac{B_1}{1-L_H},\frac{2B_1}{1-L_H}\Big)\le 0\Big).
	\end{split}\]
	Thus $u_0$ is a $\beta$-viscosity subsolution of \eqref{A1}.

	Now by Proposition \ref{dlhb}, there exists a function $u$ such that $u_0\le u \le v_0,$ $ u_*$ and $u^*$ are respectively a $\beta$-viscosity supersolution and a $\beta$-viscosity subsolution of Equation \eqref{A1} in $\Omega.$ It follows from Corollary \ref{hq3.1} that $u_*\ge u^*.$ Clearly by the definition,  $u_*\le u^*.$ Hence $u^*=u_*=u$ is a $\beta$-viscosity solution of Equation \eqref{A1} in $\Omega.$ 	\hfill $\square$
\end{proof}
\begin{example}\label{vd0T1}
	In Example 3.1, we look at
	a linear Hamiltonian $H$ and verify that assumptions (H0)-(H3) are satisfied. Here is another example, in which the Hamilton function $H$ is nonlinear. Moreover, the underlying normed space $X$ below has no equivalent Fr\'{e}chet smooth norm.

	Let $X=L_1[0,1].$ By \cite{BZ96}, $X$ has an equivalent weak Hadamard smooth norm and that norm is nowhere Fr\'{e}chet-differentiable.
	Thus, the results in Crandall \cite{CL85} do not work in this case. Also note that $L_1[0,1]$ does not satisfy the Radon-Nikodym property  (see \cite{BZ99}). Let $U=[0,1]$ and $\mathcal{U}=\{ \alpha: [0,\infty) \to U \text{ is continuous}\}$. Let a function $H: X\times X_{\beta}^*\to \mathbb{R}$ defined by
	$$H(x,p)=\sup_{\alpha \in \mathcal{U}} \{-\langle p, x_0 \alpha \rangle -|x| \alpha  \},$$
	where $x_0$ is a fixed point in $X$. Then it can be verified that $H$ satisfies conditions (H0)-(H3).
	Consider the equation
	\begin{equation}\label{TPVD}
	u+H(x,Du)=0 \quad \text{ on } X.
	\end{equation}
	By virtue of Proposition \ref{md3.2} and Theorem \ref{dl3.2}, equation \eqref{TPVD} has a unique weak Hadamard-viscosity solution.
	\end{example}

\section{Proof of Theorem \ref{dl3.1}}
\subsection{Proof for the case $\Omega\not= X$}
 The conclusion in this case is clear if $\sup_{\partial  \Omega}(u-v)^+=\infty.$ Thus, we now assume  $\sup_{\partial  \Omega}(u-v)^+<\infty.$

To continue, we first give the following lemma.
\begin{lemma}\label{bd3.1}
	Suppose that $\Omega\subset X$ is an open strict subset. We denote as $\rho(x)$ the distance from $x$ to $\partial\Omega.$ Let $u,v\in C(\overline{\Omega})$ be functions that satisfying the condition (\ref{1.10}). Then, for all $x,y \in  \Omega,$
	\begin{equation}\label{1.13}
	u(x)-v(y)\le \sup_{\partial  \Omega}(u-v)^+ + m(\min(\rho(x),\rho(y)))+m(|x-y|).
	\end{equation}
	In particularly,
	\begin{equation}\label{T3.7}
	u(x)-v(x) \le \sup_{\partial  \Omega}(u-v)^+ + m(\rho(x)), \ x\in \Omega.
	\end{equation}
\end{lemma}
\begin{proof}
	Suppose that $\rho(x)< \rho(y).$ Then, for any $\varepsilon>0,$ there exists
	an $x^*\in \partial \Omega$ such that  the line segment connecting $x$ and $x^*$ is contained in $\Omega,$ and $|x-x^*|<\rho(x)+\varepsilon.$
	For each $n\in \mathbb{N^*},$ we choose $x_n=x+\left(1-\frac{1}{n}\right)(x^*-x).$ Then,
	$x_n\in\Omega,$ and $x_n\to x^*$ as $n\to \infty.$ In addition,
	\begin{align}
	u(x)-v(y)&=[u(x)-u(x_n)]+[v(x_n)-v(x)] +[u(x_n)-v(x_n)] +[v(x)-v(y)] \notag \\
	&\le m(|x-x_n|)+u(x_n)-v(x_n)+m(|x-y|). \label{them3}
	\end{align}
	Let $n\to \infty$ in \eqref{them3}, we obtain
	\begin{equation*}
	u(x)-v(y)\le m(|x-x^*|)+u(x^*)-v(x^*)+m(|x-y|).
	\label{them4}
	\end{equation*}
	Moreover, $u(x^*)-v(x^*)\le \sup_{\partial  \Omega}(u-v)^+ $ and
	$m(|x-x^*|)\le m(\rho(x)+\varepsilon).$ Thus,
	\begin{equation}
	u(x)-v(y)\le m(\rho(x)+\varepsilon)+\sup_{\partial  \Omega}(u-v)^+ +m(|x-y|).
	\label{them5}
	\end{equation}
	The inequality \eqref{them5} holds for any $\varepsilon>0.$
	By letting $\varepsilon\to 0$ and using the continuity of the function $m$ at $\rho(x),$ we obtain
	$$ u(x)-v(y)\leq \sup_{\partial  \Omega}(u-v)^+ + m(\rho(x))+m(|x-y|). $$
	If  $\rho(x)\geq \rho(y),$ by a similar argument, we have
	\begin{equation*}
	u(x)-v(y)\le \sup_{\partial  \Omega}(u-v)^+ + m(\rho(y))+m(|x-y|).
	\label{them8}
	\end{equation*}
	Thus, \eqref{1.13} is proved.\hfill $\square$
\end{proof}

By Lemma \ref{bd3.1}, the estimate \eqref{T3.7}, and the function
$\rho(x)$ has at most linear growth, there are constants $A,B \geq 0$ such that
\begin{equation}\label{1.14}
u(x)-v(x)\le A+B|x|, \ x\in \Omega.
\end{equation}

Next, let $\zeta\in C^1(\R)$  satisfy
	\begin{equation*}
	\zeta(r)=0  \text{ with } r\le 1, \quad \zeta(r)=r-2 \text{ with }r\ge 3,\quad  \text{ and } 0\le \zeta'\le 1.
	\label{1.15}
	\end{equation*}
	For $a, \varepsilon, \lambda, R>0,$ consider the function
	\begin{equation}
	\Phi(x,y)=u(x)-v(y)-\left(\frac{|x-y|^2}{\varepsilon}+\lambda\zeta(|x|-R)\right)
	\label{1.16}
	\end{equation}
	on the set
	\begin{equation*}
	\Delta(a)=\{(x,y)\in \Omega\times \Omega: \rho(x),\rho(y)>a \text{\ \  and\ \  } |x-y|<a\}.
	\label{1.17}
	\end{equation*}
We claim that for $B$ defined in \eqref{1.14}, if
	\begin{equation*}
	\lambda >B,\quad \varepsilon\le a^2/(m(a)+1)\quad \text{  and  }\quad R>1,
	\label{1.18}
	\end{equation*}
	then
	\begin{equation}\label{1.19}
	\begin{split}
	\Phi(x,y)&\le \sup_{\partial \Omega} (u-v)^++2m(a)+C(a,\lambda,\varepsilon)\\
	\Big({\rm resp.}\quad \Phi(x,y)&\le \sup_{\partial \Omega} (u-v)^++2m(a)+ \frac{1}{1-L_H}C(a,\lambda,\varepsilon)\Big)
	\end{split}
	\end{equation}
	on $\Delta(a),$ where
	\begin{equation*}
	\begin{split}
	C(a,\lambda,&\varepsilon)=m_H(2m(a)+(\varepsilon m(a))^{1/2})+\sigma_H(\lambda,2(m(a)/\varepsilon)^{1/2}+\lambda)\\
	&+\sup\{(\widehat{H}-H)^+(z,r,p): (z,r,p)\in\Omega\times \R\times X^*\text{ and }|p|\le 2(m(a)/{\varepsilon})^{1/2}\}
	\end{split}
	\end{equation*}
	The proof of \eqref{1.19} is carried out in $2$ steps.

	\medskip
	\noindent
	\textbf{Steps 1.} Proof of \eqref{1.19} in the case $ \Phi$ attains the maximum value on $\Delta(a)$ at point $(x_0,y_0).$

	If it is the case, fix $y=y_0.$ Then the function $u(x)-\left(\frac{|x-y_0|^2}{\varepsilon}+\lambda\zeta(|x|-R)\right)$ attains a local maximum at $x=x_0.$ Hence
	$$\frac{1}{\varepsilon}(\nabla_{\beta}|.|^2) (x_0-y_0)+\lambda\zeta'(|x_0|-R).(\nabla_{\beta}|.|)(x_0)    \in D^+_{\beta}u(x_0).$$
	Similarly, when we fix $x=x_0,$ the function $v$ is $\beta$-viscosity subdifferentiable at $y_0$ and
	$$\frac{1}{\varepsilon}(\nabla_{\beta}|.|^2) (x_0-y_0)    \in D^-_{\beta}v(y_0).$$
	Since $u,v$ are $\beta$-viscosity solutions of equation  (\ref{1.9}), we have
	\begin{align}
	u(x_0)+H(x_0,u(x_0), p_{\varepsilon}+\lambda q)&\le 0,\label{1.311} \\
	v(y_0)+\widehat{H}(y_0,v(y_0),p_{\varepsilon})&\ge 0,
	\label{1.312}
	\end{align}
	where $p_{\varepsilon} =\frac{1}{\varepsilon}(\nabla_{\beta}|.|^2) (x_0-y_0), q=\zeta'(|x_0|-R).(\nabla_{\beta}|.|)(x_0).$

	Because the function  $\Phi$  attains its maximum value  at $(x_0,y_0)$  so for all $(x,y)\in \Delta(a),$ then
	\begin{equation}
	\Phi(x,y) \le \Phi(x_0,y_0)\le u(x_0)-v(y_0).
	\label{1.191}
	\end{equation}

	If $u(x_0)-v(y_0)\le 0,$ then $\Phi(x,y)\leq 0$ in $\Delta(a).$ This fact proves \eqref{1.19} because its right-hand side is nonnegative.

	If $u(x_0)-v(y_0)> 0,$ then $u(x_0)>v(y_0).$ From (\ref{1.311}) and (\ref{1.312}), we imply that
	\begin{equation}
	u(x_0)-v(y_0)\le  \widehat{H}(y_0,v(y_0),p_{\varepsilon})-H(x_0,u(x_0),p_{\varepsilon}+\lambda q).
	\label{1.33}
	\end{equation}
In addition, using (H1) (resp. (H1)*) we obtain
\begin{equation}\label{1.35}
\begin{split}
\widehat{H}&(y_0,v(y_0),p_{\varepsilon})-H(x_0,u(x_0),p_{\varepsilon}+\lambda q)\\
&= \widehat{H}(y_0,v(y_0),p_{\varepsilon})-H(y_0,v(y_0),p_{\varepsilon})+H(y_0,v(y_0),p_{\varepsilon})-H(x_0,u(x_0),p_{\varepsilon}+\lambda q) \\
&\le  \widehat{H}(y_0,v(y_0),p_{\varepsilon})-H(y_0,v(y_0),p_{\varepsilon})+H(y_0,u(x_0),p_{\varepsilon})-H(x_0,u(x_0),p_{\varepsilon}+\lambda q)\\
&= \widehat{H}(y_0,v(y_0),p_{\varepsilon})-H(y_0,v(y_0),p_{\varepsilon})
+H(y_0,u(x_0),p_{\varepsilon})-H(x_0,u(x_0),p_{\varepsilon})\\
&\ \  +H(x_0,u(x_0),p_{\varepsilon})-H(x_0,u(x_0),p_{\varepsilon}+\lambda q);\\
\Big(&{\rm resp.}\quad
\widehat{H}(y_0,v(y_0),p_{\varepsilon})-H(x_0,u(x_0),p_{\varepsilon}+\lambda q)\\
&\leq  \widehat{H}(y_0,v(y_0),p_{\varepsilon})-H(y_0,v(y_0),p_{\varepsilon})
+H(y_0,u(x_0),p_{\varepsilon})-H(x_0,u(x_0),p_{\varepsilon})\\
&\ \  +H(x_0,u(x_0),p_{\varepsilon})-H(x_0,u(x_0),p_{\varepsilon}+\lambda q)+L_H(u(x_0)-v(y_0)\Big).
\end{split}
\end{equation}

On the other hand, we have
	\begin{align*}
	0\le \Phi(x_0,y_0)- \Phi(x_0,x_0)&=v(x_0)-v(y_0)-\frac{|x_0-y_0|^2}{\varepsilon} \\
	&\le m(|x_0-y_0|)-\frac{|x_0-y_0|^2}{\varepsilon}
	\end{align*}
	so
	$$ \frac{|x_0-y_0|^2}{\varepsilon} \le  m(|x_0-y_0|) \le m(a),$$
	or $$|x_0-y_0|\le (\varepsilon m(a))^{1/2}.  $$

	Together with the assumption (H3) and the increasing monotonicity of the module $m_H,$ we have
	\begin{equation}\label{4.101}
	\begin{split}
	H(y_0,u(x_0),p_{\varepsilon})-H(x_0,u(x_0),p_{\varepsilon}) &\le m_H(2|x_0-y_0|^2/\varepsilon+|x_0-y_0|)\\
	&\le m_H(2 m(a)+(\varepsilon m(a))^{1/2}).
	\end{split}
	\end{equation}
Because of the facts that
	$$  \lambda |q|=\lambda |\zeta'(|x_0|-R).(\nabla_{\beta}|.|)(x_0)|\le \lambda,$$
	and
	$$ |p_{\varepsilon}| =\frac{1}{\varepsilon}|(\nabla_{\beta}|.|^2) (x_0-y_0) |\le \frac{2}{\varepsilon}  |x_0-y_0|\le 2 (m(a)/\varepsilon)^{1/2},$$
by using (H2) we yield
	\begin{equation}\label{4.102}
\begin{split}
H(x_0,u(x_0),p_{\varepsilon})-H(x_0,u(x_0),p_{\varepsilon}+\lambda q)&\le \sigma_H(\lambda |q|,|p_{\varepsilon}|+\lambda |q|)\\
	&\le \sigma_H(\lambda,2 (m(a)/\varepsilon)^{1/2}+\lambda).
	\end{split}
\end{equation}

Plugging \eqref{4.101}, \eqref{4.102} into \eqref{1.35}, then combining with \eqref{1.191}, \eqref{1.33} we have
\[\Phi(x,y)\leq C(a,\lambda,\varepsilon)\quad (resp.\quad  \Phi(x,y)\leq \frac{1}{1-L_H}C(a,\lambda,\varepsilon)), \]
in $\Delta(a)$ and \eqref{1.19} holds in this case.

	\medskip
	\noindent
	\textbf{Steps 2.} 	Prove (\ref{1.19}) in the case the function $\Phi$ does not attain its maximum value on $\Delta(a).$

	To do this, firstly we claim that function $\Phi$ is bounded from above on $\Delta(a).$ Indeed, since $\lambda>B,$ for any $x, y\in\Delta(a)$ we have
	\begin{align}
	\Phi(x,y)& \le u(x)-v(x)+v(x)-v(y)-\frac{|x-y|^2}{\varepsilon}-\lambda(|x|-R-2) \notag \\
	&\le A+B|x|+m(a)-\lambda|x|+\lambda(R+2)\to -\infty \text{ khi  }|x|\to \infty. \label{them2}
	\end{align}
	Then  $\Phi<0$ when $|x|>R_0$ for $R_0$ large enough. On the other hand, for $|x|\le R_0$ we have
	\begin{align*}
	\Phi(x,y)&\le u(x)-v(y)=u(x)-v(x)+v(x)-v(y)\le u(x)-v(x)+m(a)\\
	& \le A+B|x|+m(a)\le A+B R_0+m(a)<\infty
	\end{align*}
	so  $\Phi(x,y)$ is bounded from above on $\Delta(a).$

	Since $\Phi$ is bounded from above, there exists the finite $\sup_{\Delta(a)}\Phi.$

	If  $\sup_{\Delta(a)}\Phi\le\sup_{\partial \Omega} (u-v)^++2m(a),$ then (\ref{1.19}) is obvious. Otherwise, we have
	\begin{equation}
	\sup_{\Delta(a)}\Phi>\sup_{\partial \Omega} (u-v)^++2m(a).
	\label{1.20}
	\end{equation}
		From  (\ref{1.13}) and  (\ref{1.16})   we have
	\begin{equation}
	\Phi(x,y)\le  \sup_{\partial  \Omega}(u-v)^+ + m(\min(\rho(x),\rho(y)))+m(a).
	\label{1.22'}
	\end{equation}

	Choose a sequence $(x_n,y_n)\in \Delta(a), n=1,2,...$ such that $\Phi(x_n,y_n)$ is stricly increasing,
	\begin{equation}
	\Phi(x_n,y_n)\to \sup_{\Delta(a)}\Phi \text{  and  } \Phi(x_n,y_n)\ge \Phi(x_n,x_n).
	\label{1.21}
	\end{equation}
	We prove that $(x_n,y_n)$ is an interior point of $\Delta(a),$  or in other words, there exists $\gamma >0$ such that
	\begin{equation}
	S_n:=\{(x,y)\in X\times X: |x-x_n|^2+|y-y_n|^2\le \gamma^2\}\subset \Delta(a).
	\label{1.26}
	\end{equation}
	Indeed, we have
	\begin{equation}
	\Phi(x,y)-\Phi(x,x)=v(x)-v(y)-\frac{|x-y|^2}{\varepsilon}\le m(|x-y|)-\frac{|x-y|^2}{\varepsilon}
	\label{1.23'}
	\end{equation}
	on $\Delta(a).$ Substituting  $x=x_n,y=y_n$ into (\ref{1.23'}) and combining with (\ref{1.21}), we have
	\begin{equation}
	|x_n-y_n|^2\le \varepsilon m(|x_n-y_n|)\le \varepsilon m(a)\le a^2.m(a)/(m(a)+1).
	\label{1.24}
	\end{equation}
	Hence
	\begin{equation*}
	|x_n-y_n|\le a (m(a)/(m(a)+1))^{1/2}.

	\end{equation*}
	From  (\ref{1.20}) and  (\ref{1.21}) we see that there exists $\xi>0$ such that both
	\begin{equation}
	\rho(x_n) \text{ and }\rho(y_n)\text{ are not less than } a+\xi \text{ for } n \text{ sufficiently large. }
	\label{1.23}
	\end{equation}
	Indeed, otherwise, we have $\rho(x_n)\to a.$ If $x=x_n,y=y_n$ in (\ref{1.22'})  we have
	$$  \Phi(x_n,y_n)\le  \sup_{\partial  \Omega}(u-v)^+ + m(\min(\rho(x_n),\rho(y_n)))+m(a).$$
	Then let $n\to \infty$ we obtain $\sup_{\Delta(a)}\Phi\le \sup_{\partial \Omega} (u-v)^++2m(a).$
	This contradicts to (\ref{1.20}). Thus, \eqref{1.23} is proved.

	For $(x,y)\in S_n,$ we have $|x-x_n|^2+|y-y_n|^2\le \gamma^2$ so
	$$|x-x_n|+|y-y_n|\le \sqrt{2}. \left(|x-x_n|^2+|y-y_n|^2\right)^{1/2}<2\gamma$$
	and $$|x-y|\le |x-x_n|+|x_n-y_n|+|y_n-y|<2\gamma+a (m(a)/(m(a)+1))^{1/2}<a,$$
	where $\gamma$  is a positive number, which is chosen so that
	$$\gamma<\min\left(\xi, \frac{1}{2}(1-(m(a)/(m(a)+1)))^{1/2}a\right).$$
	Following \eqref{1.23} we have  $\rho(x)\ge \rho(x_n)-|x-x_n|\ge a+\xi -\gamma>a.$ A similar argument can be used to show that $\rho(y)> a,$ hence \eqref{1.26} holds.
	By \eqref{them2} the sequence  $ (x_n,y_n)$  is bounded.

	Next, we will prove that
	\begin{align}
	\sup_{\Delta(a)}\Phi  &\le C(a,\lambda,\varepsilon)\quad (resp.\quad  \sup_{\Delta(a)}\Phi\leq \frac{1}{1-L_H}C(a,\lambda,\varepsilon)), \label{them1}
	\end{align}
	 which will allow us to obtain  \eqref{1.19}.

	 Indeed, put
	\begin{equation*}
	\delta_n=\sup_{\Delta(a)}\Phi-\Phi(x_n,y_n).
	\label{1.27}
	\end{equation*}
	According to \eqref{1.21}, $\delta_n>0$ for all $n.$
	Consider the function
	\begin{equation}
	\Psi(x,y)=\Phi(x,y)-(3\delta_n/\gamma^2)(|x-x_n|^2+|y-y_n|^2) \text{ on } S_n.
	\label{1.28}
	\end{equation}
	For $(x,y)\in \partial S_n$ we have
	$$ \Psi(x,y) =\Phi(x,y)-3\delta_n\le  \sup_{\Delta(a)}\Phi-3\delta_n=\Phi(x_n,y_n)-2\delta_n=\Psi(x_n,y_n)-2\delta_n.$$
	Hence
	\begin{equation*}
	\Psi(x,y)\le \Psi(x_n,y_n)-2\delta_n\text{ on } \partial S_n.
	\label{1.29}
	\end{equation*}
	Indeed, if $P: S_n\to \R$  varies by less than $\delta_n$ on $S_n$ and $\Psi-P$ has the maximum point in $S_n,$ then this point must belong to the interior of $S_n.$

	By Proposition \ref{dlbp2}, with $\varepsilon =\delta_n, $ there  exists $P\in \mathcal{D}_{\beta}^*(X\times X)$ such that $\Psi(x,y)-P(x,y)$ attains its maximum value on $S_n$ at a point $(\widehat{x},\widehat{y})$ and  $\|P\|_{\infty}<\delta_n,$ $\|\nabla_{\beta}P\|_{\infty}< \delta_n.$
	By the argument above, it lies in the interior of  $S_n.$

	We fix  $y=\widehat{y}.$ Then we have
		\begin{multline*}
	\Psi(x,\widehat{y})-P(x,\widehat{y})= u(x)-v(\widehat{y})-\left(\frac{|x-\widehat{y}|^2}{\varepsilon}+\lambda\zeta(|x|-R)\right)\\
	-(3\delta_n/\gamma^2)(|x-x_n|^2+|\widehat{y}-y_n|^2) -P(x,\widehat{y})

	\end{multline*}
	attains its maximum on $S_n$ at the point $x=\widehat{x}.$ Hence the function $u$  is $\beta$-viscosity superdifferentiable at $\widehat{x}$ and
	\[\begin{split}
	\frac{1}{\varepsilon}(\nabla_{\beta}|.|^2) (\widehat{x}-\widehat{y})&+\lambda\zeta'(|\widehat{x}-R|)(\nabla_{\beta}|.|)(\widehat{x})\\
	&+(3\delta_n/\gamma^2)(\nabla_{\beta}|.|^2) (\widehat{x}-x_n)+\nabla_{\beta}^xP(\widehat{x},\widehat{y})    \in D^+_{\beta}u(\widehat{x}).
	\end{split}\]
	Similarly,  the function $v$ is $\beta$-viscosity subdifferentiable at $\widehat{y}$ and
	$$\frac{1}{\varepsilon}(\nabla_{\beta}|.|^2) (\widehat{x}-\widehat{y})-(3\delta_n/\gamma^2)(\nabla_{\beta}|.|^2) (\widehat{y}-y_n)-\nabla_{\beta}^yP(\widehat{x},\widehat{y})    \in D^-_{\beta}v(\widehat{y}),$$
	where $\nabla_{\beta}^x$ is the $\beta$-derivative with respect to $x$ variable.

	Since $u,v$ are respectively $\beta$-viscosity subsolution and $\beta$-viscosity supersolution of equation  (\ref{1.9}),
	\begin{align}
	u(\widehat{x})+H(\widehat{x},u(\widehat{x}), p_{1\varepsilon}+\lambda q+\theta_{1n})&\le 0,\label{1.31'} \\
	v(\widehat{y})+\widehat{H}(\widehat{y},v(\widehat{y}),p_{1\varepsilon}+\theta_{2n})&\ge 0,
	\label{1.31}
	\end{align}
	where
	\begin{equation}
	p_{1\varepsilon}=\frac{1}{\varepsilon}(\nabla_{\beta}|.|^2) (\widehat{x}-\widehat{y}),
	q=\zeta'(|\widehat{x}-R|).(\nabla_{\beta}|.|)(\widehat{x}),
	\label{1.32}
	\end{equation}
	\begin{equation*}
	\theta_{1n}=K_n.(\nabla_{\beta}|.|^2) (\widehat{x}-x_n)+\nabla_{\beta}^xP(\widehat{x},\widehat{y}),\quad \theta_{2n}=-K_n.(\nabla_{\beta}|.|^2) (\widehat{y}-y_n)-\nabla_{\beta}^yP(\widehat{x},\widehat{y}),
	\end{equation*}
	\begin{equation*}
	K_n=3\delta_n/\gamma^2.
	\end{equation*}

	Since
	$(\widehat{x},\widehat{y})\in S_n, $  the sequence $(x_n,y_n)$ is bounded, hence $(\widehat{x},\widehat{y})$ and $(x_n,y_n)$  are contained in a bounded set.

	Moreover, since  $(\nabla_{\beta}|.|^2)(x)$ is bounded by $2|x|,$ we have $$|p_{1\varepsilon}+\lambda q+\theta_{1n}|\le \frac{2}{\varepsilon}|\widehat{x}-\widehat{y}|+\lambda |\widehat{x}|+K_n |\widehat{x}-x_n|+|\nabla_{\beta}^xP(\widehat{x},\widehat{y})|$$

	and $$|p_{1\varepsilon}+\theta_{2n}|\le  \frac{2}{\varepsilon}|\widehat{x}-\widehat{y}|+2K_n |\widehat{y}-y_n|+|\nabla_{\beta}^yP(\widehat{x},\widehat{y})|.$$
	Hence $p_{1\varepsilon}+\lambda q+\theta_{1n}, p_{1\varepsilon}+\theta_{2n}$ are contained in a bounded set. Let $R_1$ be an upper bound of $\widehat{x},\widehat{y}, p_{1\varepsilon}+\lambda q+\theta_{1n}$ and $p_{1\varepsilon}+\theta_{2n}.$

	Let $R\geq R_1,$ by the continuity at $0$ of the function $w_R: X^*_{\beta}\to \mathbb{R}$ in the assumption (H0), for every $\eta>0,$ there exists a neighborhood $V_{\beta}\subset X^*_\beta$ of 0 such that $|w_R(p)|<\eta$  for all $p\in V_{\beta}.$

	Since $\tau_F$  is the strongest topology  in the class of $\tau_{\beta}$ topologies on the space $X^*,$ there  exists $\delta>0$ such that $\delta B_1\subset \frac{1}{2}V_{\beta},$ where $B_1$ is the unit ball in $X^*.$

	On account of  both $K_n$ and $\delta_n$ converge to 0 as $n\to \infty,$ we can choose $n$ large enough such that $K_n.(\nabla_{\beta}|.|^2) (\widehat{x}-x_n)\in \delta B_1;$
	$K_n.(\nabla_{\beta}|.|^2) (\widehat{y}-y_n)\in \delta B_1$ and $\delta_n<\delta.$ This implies that  $\theta_{1n}\in V_\beta$ and $ \theta_{2n}\in V_\beta.$

	Because $\Psi-P$ attains its maximum at $(\widehat{x},\widehat{y}),$ then $$\Psi(\widehat{x},\widehat{y}) \ge \Psi(x_n,y_n)+P(\widehat{x},\widehat{y})-P(x_n,y_n)=\sup_{\Delta(a)}\Phi-3\delta_n\ge 0$$
	for $n$ large enough.

	Combining with (\ref{1.16}),  (\ref{1.28}), we have
	\begin{equation}
	u(\widehat{x})-v(\widehat{y}) \ge  \Psi(\widehat{x},\widehat{y}) \ge \Psi(x_n,y_n)=\sup_{\Delta(a)}\Phi-3\delta_n\ge 0.
	\label{them7}
	\end{equation}
	Hence
	$u(\widehat{x})\ge v(\widehat{y}).$

	Since $H, \widehat{H}$ satisfy condition $(H0),$ for a sufficiently large $n,$ we have
	$$ |H(\widehat{x},u(\widehat{x}), p_{1\varepsilon}+\lambda q+\theta_{1n})-H(\widehat{x},u(\widehat{x}), p_{1\varepsilon}+\lambda q)|<|w (\theta_{1n}, R)|<\eta $$
	which implies
	$$ -H(\widehat{x},u(\widehat{x}), p_{1\varepsilon}+\lambda q+\theta_{1n})<-H(\widehat{x},u(\widehat{x}), p_{1\varepsilon}+\lambda q)+\eta  $$
	and
	$$ |\widehat{H}(\widehat{y},v(\widehat{y}),p_{1\varepsilon}+\theta_{2n})-\widehat{H}(\widehat{y},v(\widehat{y}),p_{1\varepsilon}) | <w (\theta_{1n}, R)<\eta $$
	It follows that
	$$ \widehat{H}(\widehat{y},v(\widehat{y}),p_{1\varepsilon}+\theta_{2n})< \widehat{H}(\widehat{y},v(\widehat{y}),p_{1\varepsilon})+\eta.$$
	This implies
	\begin{align}\label{bd1}
	\widehat{H}(\widehat{y},v(\widehat{y}),p_{1\varepsilon}+\theta_{2n})&-H(\widehat{x},u(\widehat{x}), p_{1\varepsilon}+\lambda q+\theta_{1n})\\
	&< \widehat{H}(\widehat{y},v(\widehat{y}),p_{1\varepsilon})-  H(\widehat{x},u(\widehat{x}), p_{1\varepsilon}+\lambda q)+2\eta.\notag
	\end{align}

	By (\ref{1.31'});  (\ref{1.31}) and (\ref{bd1}) we have
	\begin{equation*}
	u(\widehat{x})-v(\widehat{y})\le  \widehat{H}(\widehat{y},v(\widehat{y}),p_{1\varepsilon})-H(\widehat{x},u(\widehat{x}),p_{1\varepsilon}+\lambda q)+2\eta.

	\end{equation*}

	On the other hand, using (H1) (resp. (H1)*) we get
	\begin{equation}\label{1.34}
	\begin{split}
	\widehat{H}&(\widehat{y},v(\widehat{y}),p_{1\varepsilon})-H(\widehat{x},u(\widehat{x}),p_{1\varepsilon}+\lambda q)\\
	&= \widehat{H}(\widehat{y},v(\widehat{y}),p_{1\varepsilon})-H(\widehat{y},v(\widehat{y}),p_{1\varepsilon})+H(\widehat{y},v(\widehat{y}),p_{1\varepsilon})-H(\widehat{x},u(\widehat{x}),p_{1\varepsilon}+\lambda q) \\
	&\le  \widehat{H}(\widehat{y},v(\widehat{y}),p_{1\varepsilon})-H(\widehat{y},v(\widehat{y}),p_{1\varepsilon})+H(\widehat{y},u(\widehat{x}),p_{1\varepsilon})-H(\widehat{x},u(\widehat{x}),p_{1\varepsilon}+\lambda q)\\
	&=  \widehat{H}(\widehat{y},v(\widehat{y}),p_{1\varepsilon})-H(\widehat{y},v(\widehat{y}),p_{1\varepsilon})
	+H(\widehat{y},u(\widehat{x}),p_{1\varepsilon})-H(\widehat{x},u(\widehat{x}),p_{1\varepsilon})\\
	&+H(\widehat{x},u(\widehat{x}),p_{1\varepsilon})-H(\widehat{x},u(\widehat{x}),p_{1\varepsilon}+\lambda q)\\
	&\le\sup\{(\widehat{H}(z,r,p)-H(z,r,p))^+: (z,r,p)\in B_{R_0}\times \R\times X^*, |p|\le |p_{1\varepsilon}|\}\notag\\
	&+m_H(2|\widehat{x}-\widehat{y}|^2/\varepsilon+|\widehat{x}-\widehat{y}|)+\sigma_H(\lambda,2|\widehat{x}-\widehat{y}|/\varepsilon+\lambda);\\
	\Big({\rm resp.}&\quad  \widehat{H}(\widehat{y},v(\widehat{y}),p_{1\varepsilon})-H(\widehat{x},u(\widehat{x}),p_{1\varepsilon}+\lambda q)\\
	&\leq \sup\{(\widehat{H}(z,r,p)-H(z,r,p))^+: (z,r,p)\in B_{R_0}\times \R\times X^*, |p|\le |p_{1\varepsilon}|\}\\
	&+m_H(2|\widehat{x}-\widehat{y}|^2/\varepsilon+|\widehat{x}-\widehat{y}|)+\sigma_H(\lambda,2|\widehat{x}-\widehat{y}|/\varepsilon+\lambda)+L_H(u(\widehat{x})-v(\widehat{y}))	\Big),
	\end{split}
	\end{equation}
	where $B_{R_0}$  is the ball centered at 0 with  sufficiently large radius in $X.$  From \eqref{1.24} and $(x_n,y_n)\in S_n$
	we have
	\begin{equation}
	|\widehat{x}-\widehat{y}|\le |x_n-y_n|+2\gamma\le (\varepsilon m(a))^{1/2}+2\gamma.
	\label{1.36}
	\end{equation}
	It follows that
	\begin{align*}
	2|\widehat{x}-\widehat{y}|^2/\varepsilon+|\widehat{x}-\widehat{y}|&\le \frac{2}{\varepsilon}(\varepsilon m(a)+4\gamma (\varepsilon m(a))^{1/2} +4\gamma^2)+(\varepsilon m(a))^{1/2}+2\gamma\\
	&=2 m(a)+(\varepsilon m(a))^{1/2}+\gamma\left(\frac{8}{\varepsilon}(\varepsilon m(a))^{1/2}+\frac{8\gamma^2}{\varepsilon} +2\right).
	\end{align*}
	Taking $\gamma\to 0,$ we arrive at
	\begin{equation*}
	2|\widehat{x}-\widehat{y}|^2/\varepsilon+|\widehat{x}-\widehat{y}|\le 2 m(a)+(\varepsilon m(a))^{1/2}.
	\label{them6}
	\end{equation*}
	From (\ref{1.32}) and (\ref{1.36})  we have
	\begin{equation}
	|p_{i\varepsilon}|\le 2( (\varepsilon m(a))^{1/2}+\gamma)/\varepsilon.
	\label{1.37}
	\end{equation}
	From  (\ref{them7})-(\ref{1.37}) we have
	\[\Phi(x_n,y_n) -3\delta_n \le  C(a,\lambda,\varepsilon)+2\eta\]
	\[\Big({\rm resp.}\quad  \Phi(x_n,y_n) -3\delta_n \le \frac{1}{1-L_H}( C(a,\lambda,\varepsilon)+2\eta) \Big).\]

	Let $n\to \infty$ and $\eta\to0$ we have \eqref{them1}, and the inequality \eqref{1.19} is proved completely.

	Finally, choose $R\geq \max\{R_0,R_1\},$ where $R_0$ and $R_1$ appear in Step 2. By the definition of  $\zeta,$ for $|x|<R,$ $\zeta(|x|-R)=0$ and
	\begin{equation}\label{4.41}
	u(x)-v(x)=\Phi(x,x).
	\end{equation}
	Because $(x,x)\in\Delta(a),$ so from \eqref{1.19} we obtain
	\begin{equation}\label{them9}
	\begin{split}
	&u(x)-v(x)\le \sup_{\partial \Omega} (u-v)^++2m(a)+C(a,\lambda,\varepsilon)\\
	\Big({\rm resp.}&\quad u(x)-v(x)\le \sup_{\partial \Omega} (u-v)^++2m(a)+\frac{1}{1-L_H}C(a,\lambda,\varepsilon)  \Big).
	\end{split}
	\end{equation}

	Thank to \eqref{4.41},  $u-v$ is a bounded function. Thus, we can choose $B=0$ in \eqref{1.14}. By taking the limit in \eqref{them9} as $\lambda\to 0$ and $a\to 0$, we obtain \eqref{1.11}.

	\subsection{The case $\Omega=X$}
	In this case we have $\partial \Omega=\emptyset,$  so that $\sup_{\partial \Omega} (u-v)^+=0,$  $\rho(x)=\infty$ for all $x\in X.$  Using the same arguments as above, we arrive at the conclusion of the theorem. The theorem is completely proved.

\bigskip

\noindent {\bf Acknowledgements}

The authors thank an anonymous referee for careful reading
of the manuscript and her/his useful comments leading to an
improvement of the presentation.


TITLE: Does the sequence $t_n=\sqrt{a_1+\sqrt{a_2+\cdots+\sqrt{a_n}}} (a_k>0,k=1,2,\cdots)$ converge?
QUESTION [1 upvotes]: Does the sequence $t_n=\sqrt{a_1+\sqrt{a_2+\cdots+\sqrt{a_n}}} (a_k>0,k=1,2,\cdots)$ converge?
I need to prove that if
$$ \limsup\limits_{n\to\infty} \dfrac{\ln\ln a_n}{n}<\ln 2, $$
the sequence $\{t_n\}$ converge. (When $a_n\le 1$, $\ln\ln a_n$ is defined to be $-\infty$.)
I have no idea about this question. But if I consider the sequence
$$ p_n^{(r)} = \sqrt[r]{a_1+\sqrt[r]{a_2+\cdots+\sqrt[r]{a_n}}}, r\in\mathbb{Z}_+$$
and in particular, when $r=1$, I have that
$$ p_n^{(1)} = \sum_{k=1}^n a_k, $$
thus, if
$$ \limsup\limits_{n\to\infty}\dfrac{\ln a_n}{n}<0(=\ln 1), $$
it's easy to obtain that $\{p_n^{(1)}\}$ converges. I guess there is a connection but I don't know how to  continue.
Any hint?

REPLY [0 votes]: This is equivalent to showing that there exists a constant $R$ such that $a_n\leq R^{2^n}$. To see this, first unwrap the $\sup$ condition you're given for large enough $n$ via:
$$\ln(\ln(a_n))/n\leq \ln(2)+\epsilon$$
for all $n\geq N$, giving $a_n\leq ce^{2^n}\leq R^{2^n}$.
Thus:
\begin{align*}
\sqrt{a_1+\sqrt{a_2+\cdots+\sqrt{a_n}}}&\leq \sqrt{R^{2^1}+\sqrt{R^{2^2}+\cdots+\sqrt{R^{2^n}}}}\\
&= R\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}\\
&\leq 2R,
\end{align*}
and since the sequence is monotone increasing and bounded, we're done.


TITLE: Calculate Fourier Transform of $\frac{\sin^2(x)}{x^2}$ for $k\in[0,2]$
QUESTION [2 upvotes]: So here we have that the Fourier Transform equals:
$$\int_{-\infty}^{\infty}\frac{\sin^2(x)}{x^2}e^{-ikx}dx=-\frac{1}{4}\int_{-\infty}^{\infty}\frac{-2e^{-ikx}+e^{i(2-k)x}+e^{-i(2+k)x}}{x^2}dx$$.
Now since $k\in[0,2]$, for contour integration, we consider the LHP for the first and third term in the integrand and then the UHP for the second term.
If we call $C_{\epsilon,U}$ the semi-circle of radius $\epsilon$ in the UHP, $C_{\epsilon,L}$ the semi-circle of radius $\epsilon$ in the LHP, $C_{R,U}$ the semi-circle of radius $R$ in the UHP, $C_{R,L}$ the semi-circle of radius $R$ in the UHP, with closed contours $\Gamma_{U}$ and $\Gamma_{L}$, we have:
$$\int_{-\infty}^{\infty}=\lim_{R\to\infty}\lim_{\epsilon\to 0}\Bigg(\int_{\Gamma_U}-\int_{C_{\epsilon,U}} -\int_{C_{R,U}}\Bigg) \text{ for the second term}$$
$$\int_{-\infty}^{\infty}=\lim_{R\to\infty}\lim_{\epsilon\to 0}\Bigg(-\int_{\Gamma_L}-\int_{C_{\epsilon,L}} -\int_{C_{R,L}}\Bigg) \text{ for the remaining terms}$$
(Note the negative sign in line 2 due to negative orientation of $\Gamma_L$)
No poles lie inside the closed $\Gamma_U$ and $\Gamma_L$ so the first integral on the RHS of each line is zero, and the last integral on the RHS each line $\to 0$ as $R\to \infty$.
So, for the top line, we're left with:
$$ -\frac{1}{4}\int_{-\infty}^{\infty}\frac{e^{i(2-k)x}}{x^2}dx=\frac{1}{4}\lim_{\epsilon\to 0}\int_{C_{\epsilon,U}}\frac{e^{i(2-k)z}}{z^2}dz=\frac{1}{4}\lim_{\epsilon\to 0}\int_{\pi}^{0}\frac{e^{i(2-k)\epsilon e^{i\theta}}}{\epsilon^2e^{2i\theta}}i\epsilon e^{i\theta}d\theta$$.
Expanding $e^{i(2-k)\epsilon e^{i\theta}}=1+i(2-k)\epsilon e^{i\theta}+...$ (can ignore higher order terms as they tend to $0$ as $\epsilon \to 0$).
So we get (for the second term in the original integrand): $$-\frac{1}{4}\lim_{\epsilon\to 0}\int_{0}^{\pi}\frac{1+i(2-k)\epsilon e^{i\theta}}{\epsilon^2e^{2i\theta}}i\epsilon e^{i\theta}d\theta=-\frac{1}{4}\lim_{\epsilon\to 0}\int_{0}^{\pi}\frac{ie^{-i\theta}}{\epsilon}d\theta+\frac{1}{4}\int_{0}^{\pi}(2-k)d\theta=-\frac{1}{4}\lim_{\epsilon\to 0}\frac{2}{\epsilon}+\frac{1}{4}(2-k)\pi$$
Now when we consider the remaining terms (the 1st and 3rd terms in the original integrand, using the LHP), using the same method I get:
$$-\frac{1}{4}\int_{-\infty}^{\infty}\frac{-2e^{-ikx}+e^{-i(2+k)x}}{x^2}dx$$
$$=\frac{1}{4}\lim_{\epsilon\to 0}\int_{-\pi}^{0}\frac{-2(1-i\epsilon e^{i\theta}k+...)+(1-i\epsilon e^{i\theta}(2+k)+...)}{\epsilon^2e^{2i\theta}}i\epsilon e^{i\theta}d\theta$$ $$=\frac{1}{4}\lim_{\epsilon\to 0}\int_{-\pi}^{0}-\frac{ie^{-i\theta}}{\epsilon}d\theta+\frac{1}{4}\int_{-\pi}^{0}(2-k)d\theta= \frac{1}{4}\lim_{\epsilon\to 0}\frac{2}{\epsilon}+\frac{1}{4}(2-k)\pi$$
So adding both integrals together, I get that the Fourier transform is $\pi(1-\frac{1}{2}k)$. Is this correct? I feel like I may have gone wrong somewhere?

REPLY [3 votes]: BRUTE FORCE METHODOLOGY:
Not quite.  There is a subtle error in the rigor applied.  The limiting operation (as $\varepsilon\to 0$) cannot be separated.
So, to proceed rigorously, we denote the integral of interest as $F(k)$ and write
$$\begin{align}
F(k)&=\int_{-\infty}^\infty \frac{\sin^2(x)}{x^2}e^{-ikx}\,dx\\\\
&=\frac14\int_{-\infty}^\infty \frac{2e^{-ikx}-e^{-i(k-2)x}-e^{-i(k+2)x}}{x^2}\,dx\\\\
&=\frac14 \lim_{\varepsilon\to 0^+}\left(\int_{|x|\ge \varepsilon} \frac{2e^{-ikx}-e^{-i(k+2)x}}{x^2}\,dx- \int_{|x|\ge \varepsilon} \frac{e^{-i(k-2)x}}{x^2}\,dx\right)\tag1
\end{align}$$
Now we analyze the following closed contour integrals, $I_1$ and $I_2$, with $0<\varepsilon<R$, as given by
$$\begin{align}
I_1&=\int_{R\ge |x|\ge \varepsilon} \frac{2e^{-ikx}-e^{-i(k+2)x}}{x^2}\,dx\\\\
&+\int_{\pi}^{2\pi}\frac{2e^{-ik\varepsilon e^{i\phi}}-e^{-i(k+2)\varepsilon e^{i\phi}}}{(\varepsilon e^{i\phi})^2}\,i\varepsilon e^{i\phi}\,d\phi\\\\
&+\int_{2\pi}^{\pi}\frac{2e^{-ikR e^{i\phi}}-e^{-i(k+2)R e^{i\phi}}}{(R e^{i\phi})^2}\,iR e^{i\phi}\,d\phi\tag2
\end{align}$$
and
$$\begin{align}
I_2&=\int_{R\ge |x|\ge \varepsilon} \frac{e^{-i(k-2)x}}{x^2}\,dx\\\\
&+\int_{\pi}^{0}\frac{e^{-i(k-2)\varepsilon e^{i\phi}}}{(\varepsilon e^{i\phi})^2}\,i\varepsilon e^{i\phi}\,d\phi\\\\
&+\int_{0}^{\pi}\frac{e^{-i(k-2)R e^{i\phi}}}{(R e^{i\phi})^2}\,iR e^{i\phi}\,d\phi\tag3
\end{align}$$
As $R\to \infty$, the third integrals on the right-hand sides of $(2)$ and $(3)$ vanish.  Application of Cauchy's Integral Theorem guarantees that
$$\begin{align}
F(k)&=\frac14 \lim_{\varepsilon \to 0^+}\left(-\int_{0}^{\pi}\frac{e^{-i(k-2)\varepsilon e^{i\phi}}}{(\varepsilon e^{i\phi})^2}\,i\varepsilon e^{i\phi}\,d\phi-\int_{\pi}^{2\pi}\frac{2e^{-ik\varepsilon e^{i\phi}}-e^{-i(k+2)\varepsilon e^{i\phi}}}{(\varepsilon e^{i\phi})^2}\,i\varepsilon e^{i\phi}\,d\phi\right)\\\\
&=\pi(1-k/2)
\end{align}$$
as expected.

ALTERNATIVE METHODOLOGY:
I thought it might be instructive to present an alternative, simpler solution.  We begin by integrating by parts the original integral with $u=2e^{-ikx}-e^{-i(k-2)x}-e^{-i(k+2)x}$ and $dv=-\frac1{4x}$.  Proceeding, we find that
$$\begin{align}
F(k)&=-\int_{-\infty}^\infty \frac{-i2k e^{-ikx}+i(k-2)e^{-i(k-2)x}+i(k+2)e^{-i(k+2)x}}{4x}\,dx\\\\
&=-\frac\pi 2 k\,\text{sgn}(k)+\frac\pi4 (k-2)\,\text{sgn}(k-2)+\frac\pi4 (k+2)\,\text{sgn}(k+2)\\\\
&=\begin{cases}
0&,|k|>2\\\\
\pi(1-|k|/2)&,0\le |k|\le 2\\\\
\end{cases}
\end{align}$$


text:
\section{Conclusions and open problems}\label{chpt:future}

In this work, we study two classes of graphs: \fpExt{k} and \fppExt{k} graphs, subclasses of $k$-thin and proper $k$-thin graphs, respectively.
Concerning \fpExt{k} graphs, we present a polynomial time algorithm that receives as input a graph $G$ and a $k$-partition of $V(G)$ and decides whether $G$ is a precedence $k$-thin graph with respect to the given partition. This result is presented in Section~\ref{chpt:pthinness}. Regarding \fppExt{k} graphs, for the same input, we prove that if $k$ is a fixed value, then it is possible to decide whether $G$ is a precedence proper $k$-thin graph with respect to the given partition in polynomial time. For variable $k$, the related recognition problem is \NP-complete. These results are presented in Section~\ref{chpt:ppthinness}. Also, using threshold graphs, we characterize both \fpExt{k} and \fppExt{k} graphs.

Concerning the classes defined in this paper, some open questions are highlighted:

\begin{itemize}

    \item Given a graph $G$, what is the complexity of evaluating  \fpPar{G} and \fppPar{G}?

    \item Given a graph $G$ and an integer $k$, what is the complexity of determining  if \fpPar{G}, or \fppPar{G}, is at most $k$?

    \item How do \fpPar{G} and \fppPar{G} relate to $thin(G)$ and $pthin(G)$, respectively?

    \item Is it possible to extend the results of this paper to consider other types of orderings (partial orders) and restrictions?

\end{itemize}

config: arxiv


TITLE: Transition time in a Lotka-Volterra system
QUESTION [5 upvotes]: I am working with a set of real-valued ordinary differential equations based on the Lotka-Volterra competition equations:
$$\begin{align}
\dot{a_1} & = a_1 \left( 1 - a_1 - 2 a_2 \right) \\
\dot{a_2} & = a_2 \left( 1 - a_2 - (1-1/\nu) a_1 \right)
\end{align}$$
where $a_{1,2} \in [0,1]$ and $\nu \ge 1$. I would like to obtain a closed form (or analytical) solution for the time, $\tau$, it takes for this system to transit between two regions in state space. Specifically, I would like to solve for $\tau$ given $a_1(0) = \delta$ and $a_2(\tau) = \delta$, where $1/2 \lt \delta \lt 1$. This is along the manifold between two fixed points: the unstable manifold of a saddle at $(a_1,a_2) = (1,0)$ and a stable equilibrium at $(a_1,a_2) = (0,1)$.
There is a solution for $\nu = 1$ ($\tau = -2 \text{ln}(1/\delta-1)$), but I have been unable to find a solution for the more general case in terms of $\nu$ (or even for any particular value of $\nu \gt 1$, including the limiting case of $\nu \rightarrow \infty$). This paper seems to imply that the equations above are not fully integrable except when $\nu = 1$ (see Eq. 25'), i.e, when the second equation is not a function of $a_1$. However, I'm not actually interested in solving this system for all time over the full state space. Question: Are there any methods to solve or obtain a reliable approximation for $\tau = f(\nu, \delta)$ for this system just within my region of interest?
Attempts:

In addition to paper and pencil, I have used Matlab's dsolve and Mathematica's DSolve along with assumptions to try to solve the ODEs for the specified boundary conditions. I was unable solve the system using these, but might there be ways to transform or break up the system that would facilitate a solution?
I have tried using low-order power series, e.g., this paper, about each of the equilibria to obtain functions of time that are then inverted to solve for $\tau$. This was far from accurate as only a few series terms can be used.
I have tried schemes based on simulating the ODE and fitting the transit times to a function of $\nu$. This requires finding initial conditions that lie on the manifold. How can that be done reliably as a function of $\nu$ and $\delta$? And can fitting methods be adapted to more general forms of the equation above (i.e., reusing the same fitting coefficients and without fitting high dimensional surfaces)? I am more interested in non-fitting-based solutions, but if you can demonstrate something that works well, I would be happy to look at it.

Update 1 – Mar. 17, 2014:
The nullclines of the system and the Jacobian determinant (related to curvature) can be used to obtain estimated initial and final conditions. Solving for the roots of $\det(J)=0$ as a function of $a_1$ and scaling appropriately, one obtains
$$ a_2(t) = \tfrac{1}{2} \left( 2 - a_1(t) - \sqrt{a_1(t) \left( \tfrac{2}{\nu} \left( a_1(t)-1 \right) + 2 - a_1(t) \right)} \right)$$
For $a_1(0) = 1-\delta$, this expression appears to provide a good approximation for $a_2(0)$ on the stable manifold (attracting contour) in question. It is less reliable for obtaining an estimate for $a_1(\tau)$. Perhaps this is due to the Jacobian evaluated at $(a_1,a_2) = (0,1)$ being a defective matrix with only one eigenvector.
Update 2 – Mar. 25, 2014 – (post bounty):
The equation from @JJacquelin's answer immediately after the transformation to polar coordinates can be simplified and integrated definitely (I used Mathematica 9) with respect to the bounds at times $0$ and $\tau$:
$$\begin{align}
\int_0^\tau{dt} & = \int_{\theta_0}^{\theta_{\tau}}{\frac{\sin^3\theta + \cos^3\theta + \left(2\cos\theta + \left(1-\tfrac{1}{\nu}\right)\sin\theta\right)\sin\theta\cos\theta}{\left(\tfrac{1}{\nu} \cos \theta + \sin \theta\right)\sin\theta\cos \theta}d\theta} + \int_{\rho_0}^{\rho_{\tau}}{\frac{1}{\rho}d\rho} \\
\tau & = \int_{\theta_0}^{\theta_{\tau}}{\frac{1-\tfrac{1}{\nu}+\left(2+\cot\theta\right)\cot\theta+\tan\theta}{1+\tfrac{1}{\nu}\cot\theta}d\theta} + \ln\left(\frac{\rho_{\tau}}{\rho_0}\right) \\
& = \frac{\nu^2}{1+\nu^2}\left(\ln\left(\frac{\cos(\theta_0)}{\cos(\theta_{\tau})}\right) +2\ln\left(\frac{\cos\theta_{\tau}+\nu\sin\theta_{\tau}}{cos\theta_0+\nu\sin\theta_0}\right)+\frac{1}{\nu^2}\ln\left(\frac{\cos\theta_{\tau}\left(1+\nu\tan\theta_{\tau}\right)^2}{\cos\theta_0\left(1+\nu\tan\theta_0\right)^2}\right)\right) + \nu\ln\left(\frac{\nu+\cot\theta_0}{\nu+\cot\theta_{\tau}}\right) + \ln\left(\frac{\rho_{\tau}}{\rho_0}\right)
\end{align}$$
Transforming back from polar to Cartesian coordinates further simplifies the expression:
$$\tau = \ln\left(\frac{a_1(0)}{a_1(\tau)}\right) + 2\ln\left(\frac{a_1(\tau)+\nu a_2(\tau)}{a_1(0)+\nu a_2(0)}\right) + \nu\ln\left(\frac{a_2(\tau)\left(a_1(0)+\nu a_2(0)\right)}{a_2(0)\left(a_1(\tau)+\nu a_2(\tau)\right)}\right)$$
Here is some "simple" Matlab code that demonstrates this. Also in the code is an ODE-based method that uses ode45's event detection. This accurately finds the time $\tau$ by integrating along the manifold in question and terminating when the solution satisfies a condition. It's simple and fast for this basic case. However, recall that this system is a simplified version of a more general one in which the time scaling of the vector field can vary and even be different in each of the two dimensions. This could lead to long integration times if the scaling cannot be factored. I'm hoping that the analytical solution given here can be generalized.
The analytic solution-based part of my Matlab code underestimates $\tau$ (except for $\nu$ close to $1$ when it slightly overestimates it). The source of almost all of the error is due to the estimate of $a_1(\tau)$ (af1 in the code). Something else I'm looking into.

REPLY [3 votes]: We first define rigorously the set $S$ of interest, named "the stable manifold" in the question. To do so, note that the nonnegative quadrant $x\geqslant0$, $y\geqslant0$, is stable by the dynamics, and that, for every starting point in the positive quadrant $x\gt0$, $y\gt0$, the dynamics converges to $(0,1)$. Consider the reversed dynamics, defined as
$$
x'=-x\cdot(1-x-2y),\qquad y'=-y\cdot(1-y-(1-1/\nu)x).
$$
Every solution of the reversed dynamics starting from $(x_0,y_0)$ in the strip $x\in(0,1)$, $y\gt0$, either reaches the line $x=1$ at some point $(1,y)$ with $y\gt0$, or at $(1,0)$, or converges to $(0,0)$.
The "stable manifold" from $(1,0)$ to $(0,1)$ for the original dynamics is the set $S$ of initial conditions $x_0\in(0,1)$, $y_0\gt0$, such that the solution of the reversed dynamics reaches $x=1$ at $(1,0)$. Very loosely speaking, $S$ can be viewed as the trajectory of the original dynamics starting from $(1,0)$.

Second we revisit the case $\nu=1$. Then, $S$ is the line $a_1+a_2=1$ hence $\tau$ is the time to go from $(a_1,a_2)=(\delta,1-\delta)$ to $(a_1,a_2)=(1-\delta,\delta)$. The differential equation $a_2'=a_2(1-a_2)$ has an explicit solution, namely,
$$
\frac{a_2(t)}{1-a_2(t)}=\mathrm e^t\cdot\frac{a_2(0)}{1-a_2(0)}.
$$
Plugging in this identity the values $a_2(0)=1-\delta$ and $a_2(\tau)=\delta$ yields
$$
\tau=2\log\left(\frac{\delta}{1-\delta}\right),
$$
that is, the formula in the post (with a minus sign missing). More generally, the time to go from $a_2(0)$ to $a_2(\tau)\geqslant a_2(0)$ is
$$
\tau=\log\left(\frac{a_2(\tau)\cdot(1-a_2(0))}{(1-a_2(\tau))\cdot a_2(0)}\right),
$$
hence, when $a_2(0)\to0$ and $a_2(\tau)\to1$,
$$
\tau\sim-\log a_2(0)-\log(1-a_2(\tau)).
$$

Consider finally $\nu\gt1$. Let $\mu=1/\nu$. On the line $(1-\mu)a_1+a_2=1$, $a_2'=0$ and $a'_1\lt0$. On the line $a_1+2a_2=1$, $a_1'=0$ and $a'_2\gt0$. Thus the region between these two lines is stable and $S$ lies entirely between these two lines. This implies that, everywhere on $S$,
$$
\mu a_2(1-a_2)\leqslant a_2'\leqslant a_2(1-a_2).
$$
Considering the differential system near the point $(a_1,a_2)=(1,0)$, one sees that, when $\delta\to1$, the point on $S$ such that $a_1=\delta$ is such that $a_2\sim(1-\delta)^\mu$, thus the time $\tau$ to go from $a_1=\delta$ to $a_2=\delta$ is roughly the time to go from $a_2=(1-\delta)^\mu$ to $a_2=\delta$. The exact result in the case $\nu=1$ and the double inequality on $a_2'$ yield
$$
\liminf_{\delta\to1}\left(\frac{\tau}{-\log(1-\delta)}\right)\geqslant\mu(1+\mu),
$$
and
$$
\limsup_{\delta\to1}\left(\frac{\tau}{-\log(1-\delta)}\right)\leqslant1+\mu.
$$
Loosely speaking, for some $\mu(1+\mu)\leqslant\alpha\leqslant1+\mu$,
$$
\mathrm e^\tau\approx(1-\delta)^{-\alpha}.
$$
Since $\alpha\lt2$, this is a speed-up when compared to the $\nu=1$ case, due to the fact that starting from $a_1(0)=\delta$ on $S$ is to start from $a_2(0)\gg1-\delta$.


TITLE: Convergence of $\sum \frac{b_n}{n}$ Where $b_n = 1, -1, -1, 1, 1, 1, -1, -1,-1,-1,1,1,1,1,1,....$
QUESTION [8 upvotes]: How can I show convergence or divergence of the following sequence?
$$\sum \frac{b_n}{n}$$
Where $b_n = 1, -1, -1, 1, 1, 1, -1, -1,-1,-1,1,1,1,1,1,....$
Im not sure how to use any of the standart theorems to show convergence or divergence.
Any attempts to use comparison test, or finding an upper bound failed.
Any ideas?

REPLY [1 votes]: Let $S_k=b_1+b_2+\ldots +b_k.$
Denote $$u_n={2n(2n-1)\over 2}=n(2n-1)\quad  v_n={(2n+1)2n\over 2}=(2n+1)n$$ Then
$$S_{u_n}=n,\qquad S_{v_n}=-1$$
Moreover $S_k$ is decreasing for $u_n\le k\le v_n$ and increasing for $v_n\le k\le u_{n+1}.$
Therefore
$$|S_k|\le n+1,\qquad u_n\le k\le u_{n+1}$$
The condition $k\ge u_n$ implies $n\le \sqrt{2k}$ hence$$|S_k|\le \sqrt{2k}+1$$
We obtain (setting $S_0=0$)
$$\sum_{k=1}^N{b_k\over k}=\sum_{k=1}^N{S_k-S_{k-1}\over k}= \sum_{k=1}^{N-1}{S_k\over k(k+1)}+{S_N\over N} $$
The right hand side is convergent when $N\to \infty.$ So the series is convergent. Moreover we can estimate the rate of convergence
$$\left |\sum_{n=N}^\infty {b_n\over n}\right |\le {12\over \sqrt{N}}$$


TITLE: Flatten directed, acyclic graph
QUESTION [0 upvotes]: Suppose I have a directed, acyclic graph where $A \to B$ represents that $A$ depends on $B$.
I want to 'flatten' this graph so that each vertex has an edge to all direct and indirect dependencies. Ie. create a new graph such that if $A \to B$ and $B \to C$ in the original graph, then the new graph will contain $A \to B$, $B \to C$, and $A \to C$.
What is the proper name for this operation, and are there existing algorithms that can perform this task?

REPLY [1 votes]: This is the transitive closure graph (or sometimes reachability graph) and can be implemented in nearly all graph software.  In Mathematica:
g = Graph[{1 \[DirectedEdge] 2, 1 \[DirectedEdge] 3,
   1 \[DirectedEdge] 4, 3 \[DirectedEdge] 4, 2 \[DirectedEdge] 4,
   4 \[DirectedEdge] 5}, VertexLabels -> "Name"]

TransitiveClosureGraph[g, VertexLabels -> "Name"]


TITLE: $3$ children riddle, compute the ages based on information given
QUESTION [1 upvotes]: A man has $3$ children such that their ages add up to some number $x$, and whose ages multiply to some number $y$, such that $xy = 756$.  What are the ages of the $3$ children?
Letting the ages be $a$, $b$, and $c$ of the three children, what we know is the following.
$$a+b+c = x$$
$$abc = y$$
$$xy = 756.$$
How can I go about solving this?  I tried just plugging in some numbers and can get semi close such as ages $3,3,7$ which gives an $xy$ value of $819$.
Also I tried working backwards from $756$ to divide thru  by factors and I got $378,189,63,21,7$, which is why I thought one of the ages might be $7$.

REPLY [0 votes]: There is also another "reasonable" solution which is the ages are $3.5$, $4$, and $4.5$.  Some children's ages are expressed in "halves" although it is not as common as "wholes".  ($3.5 + 4 + 4.5$) * ($3.5 * 4 * 4.5$) does indeed equal $756$.  It is interesting to note that this is $12 * 63$ and if you were to guess the ages were $4,4,4$, that would give you $12 * 64$ which is very close to the correct answer of $12 * 63$ so that would imply that the ages have to all be very close to $4$ meaning they are likely not all integers.  Ages $3,4,5$ is also a near solution at $12*60=720$.


TITLE: Two-sample test for ordinal data.
QUESTION [3 upvotes]: I have a question in a survey X that can be rated between 1 and 10 (ordinal). The answers can be split in group A and group B.
I want to know if the mean of group A's answers significantly differ from groups B rating. Which test is the best one to do so and how can I do this with SPSS?
Thank you very much for your help!

REPLY [1 votes]: This seems to be a two-sample test with Groups 1 (of size $n_1$) and 2 (of size $n_2$). Your data are scores from 1 to 10 on the question.
Welch t test. If $n_1$ and $n_2$ are large enough (perhaps both above 20), you might be
able to get a reliable answer using a Welch 2-sample t-test.
Wilcoxon test. You are almost sure to have lots of ties (repeated scores) even if both
sample sizes are relatively small. Thus you will get error messages
about ties when trying to do a Wilcoxon rank-sum test, along with an
approximated P-value or a statement that a P-value is not available (depending
of the software you use).
Permutation test. Perhaps it is best to do a permutation test. Under the null
hypothesis that the two groups tend to give the same responses to the
question, the argument is that the scores could be permuted between
Groups A and B without effect. So if we choose some measure of difference
such as the difference $D = \bar X_1 - \bar X_2$ between the two sample means,
we can use either combinatorics or simulation to get the null permutation
distribution of $D$, and judge whether your observed value of $D$ is
consistent with the null distribution.
Example. I will illustrate each kind of test using fake data with 25
subjects in each group (although none of the tests require sample sizes
to be equal).
Here are listings and summaries of some fake data to use for testing.
x1; summary(x1)
##  9  6  4 10  5  5  8  8  8  8  8  9  8  6  4  7  8  9  8  6  8  8  5  8  9
##  Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
##  4.00    6.00    8.00    7.28    8.00   10.00

x2; summary(x2)
## 10  9 10  7  7  8  8 10  8  5  7  7  7  5  8 10 10 10  9  7  9 10 10 10  9
##  Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
##   5.0     7.0     9.0     8.4    10.0    10.0

A quick look shows means to be greater in Group 2 than in Group 1.
Is this difference statistically significant?
t test: A Welch 2-sample t test in
R statistical software finds a significant difference. (P-value $\approx$ 2%.) The only doubt
is whether data are sufficiently nearly normally distributed for the
t test to give accurate results. (Data for both groups spectacularly fail
a Shapiro-Wilk test with P-values < .01. But sample sizes may be large enough for the t test to be useful anyhow.)
t.test(x1, x2)

##        Welch Two Sample t-test

## data:  x1 and x2
## t = -2.434, df = 47.853, p-value = 0.01872
## alternative hypothesis: true difference in means is not equal to 0
## sample estimates:
## mean of x mean of y
##      7.28      8.40

Wilcoxon test: The Wilcoxon test, for a difference in medians gives a (tentative) P-value of about 2%, but warns that it may not be accurate.
However, there are only seven uniquely different values among the 50 subjects.
So the number of ties is 'massive' and the Wilcoxon test is based on
a comparison of ranks, which can be problematic when there are many ties. I would not want to trust the
result of the Wilcoxon test.
wilcox.test(x1, x2)

##     Wilcoxon rank sum test with continuity correction

## data:  x1 and x2
## W = 200.5, p-value = 0.02702
## alternative hypothesis: true location shift is not equal to 0

## Warning message:
## In wilcox.test.default(x1, x2) : cannot compute exact p-value with ties

Permutation test. It would be tedious to derive the exact permutation distribution of $D$ for this example. The usual cure is to simulate
a large number of permutations and to approximate the P-value from
simulation results. Here is a brief program in R to find the approximate
P-value (2.1%) of the permutation test. (You may get a slightly
different P-value at each run of the program, but not enough different
to matter in the interpretation. For this program, subsequent
runs all gave values rounding to 2%)
m = 10^4;  d.perm = numeric(m)
all = c(x1, x2);  d.obs = mean(x1) - mean(x2)
n1 = n2 = 25
for (i in 1:m) {
  perm = sample(all, n1+n2)
  d.perm[i] = mean(perm[1:n1]) - mean(perm[(n1+1):(n1+n2)])
  }
mean(abs(d.perm) >= abs(d.obs))
## 0.0215

Here is a histogram of the approximate permutation distribution. The
solid red line at the left is the observed value of $D$ for the data above.
The dotted red line at the right is just as extreme (far from 0) as the
observed value of $D.$ The P-value of this 2-sided permutation test
is the percentage of values in the permutation distribution outside these
red lines, in this case, 2.1%.

Conclusion: The two groups differ significantly. The t test is probably OK,
because, for samples this large, the Central Limit Theorem tends to make the sample means very
nearly normal even if the data are not normal. For groups as small as ten,
I would certainly insist on seeing permutation test results before drawing
a conclusion.
You can read more about permutation tests in this paper by Eudey. The two-sample test
above is discussed, with additional examples, in Section 4.
Almost certainly, your data will look different than my fake data. Please let
me know if you have trouble relating my answer to your specific data.
Note: The fake data above were generated from populations with respective
means about 3/5 and 5/6 using the R code below. (So it is appropriate that the tests found a significant difference.) By using the same seed
I used, you should get exactly the same data.
set.seed(1234)
x1 = ceiling(10*rbeta(25, 3, 2))
x2 = ceiling(10*rbeta(25, 5, 1))

Addendum (Your Data from Comment).  Your result in the Comment seems OK.
Significant at 9.3% < 10% level; sometimes optimistically called "suggestive" of significance.
If you honestly expected (before seeing data) Gp2 scores to be
higher, then maybe this should be a left-sided test of $H_0: \mu_1 \ge \mu_2$ vs.
$H_a: \mu_1 < \mu_2.$ if so, P-value would be
3.8% < 5% for significance at the 5% level.
x1 <- c(0,7,10,0,9,5,10,6,8,7,8,2,2,8,10,7,10)
x2 <- c(7,4,10,10,9,10,10,9,10,7,5,10,10,10,10,5,10,2)
all = c(x1, x2);  gp = rep(0:1; times = c(17,18))
stripchart(all~gp, meth="stack", pch=19, col=c("blue", "green3"))

Welch t-test gives P-value  0.09024.  Repeat of permutation test with
m = 10^6 iterations to reduce possibility of simulation error.
x1 <- c(0,7,10,0,9,5,10,6,8,7,8,2,2,8,10,7,10)
x2 <- c(7,4,10,10,9,10,10,9,10,7,5,10,10,10,10,5,10,2)
m = 10^6;  d.perm = numeric(m)
all = c(x1, x2);  d.obs = mean(x1) - mean(x2)
n1 = length(x1);  n2 = length(x2)
for (i in 1:m) { perm = sample(all, n1+n2)
  d.perm[i] = mean(perm[1:n1]) - mean(perm[(n1+1):(n1+n2)]) }
mean(abs(d.perm) >= abs(d.obs))
## 0.093149
## 0.093183  # 2nd run with m=10^6

mean(d.perm < d.obs)
## 0.038349  # P-value of LEFT SIDED test

length(unique(d.perm))
## 75        # uniquely different sim. values of D (enough)

hist(d.perm, prob=T, col="skyblue2", main="Simulated Permutation Distribution")
abline(v=d.obs, col="red", lwd=2)
abline(v=-d.obs, col="red", lwd=2, lty="dashed")

Note: If this is for a reviewed paper, you might get criticism
(as noted by @Nameless) that the permutation test involves taking
sample means of ordinal data. Possible nonparametric, ordinal-oriented alternatives:
(a) Use
median instead of mean in the permutation test when finding
d.obs and (within the loop) when finding d.perm, but not at
the end when finding the P-value. (In R, the mean of a logical vector
is the proportion of its TRUEs.) Trouble is I got only only about 20
uniquely different values of d.perm that way; not quite enough
for my taste. One-sided P-value 0.047.
(b) Do a Welch t test
on rank-transformed data. (Ranks are appropriate for ordinal data, their means are
likely not far from normal with sample sizes above 15.) From
t.test(rank(all) ~ gp, alte="less"), I get (Welch, one-sided) P-value 0.03457.


TITLE: Rigorous proof of the duality of Coupon collector's problem and Occupancy problem
QUESTION [0 upvotes]: We have $k$ different types of coupons (with replacement).If we collect at least $l$ different coupons, we win a prize. We can only afford to collect $m$ coupons.
Let's say we take all those $m$ coupons, and we collect exactly $L$ different coupons. Then the probability that we win a prize is $P(L \geq l)$
Let $T$ be the number of trials taken to collect exactly $l$ coupons. Then the probability we win a prize is $P(T \leq m)$
"Clearly" $P(L \geq l) = P(T \leq m)$.
So how on earth does one go about proving it rigorously? I'm not too familiar with the theory of stopping times, and the proof may come from that area, but no treatise that mentions these two problems ever seems to prove this equality. Does anyone know of someone that has taken the time to prove this rigorously, and what formalism that they use?

REPLY [6 votes]: To specify the probability space is rarely a good idea but since you insist on it, a possible choice would be $\Omega=K^\mathbb N$, where $K=\{1,2,\ldots,k\}$ denotes the set of types of coupons, endowed with the sigma-algebra ${\mathcal K}=2^K$ and the discrete measure $p=(p_i)$ on $(K,\mathcal K)$ describing the probability to get a coupon of a given type at each draw. In other words, for every $A\subseteq K$, $A\in\mathcal K$ and $p(A)=\sum\limits_{i\in A}p_i$.
This considers that one collects coupons again and again, even when one is really interested in what happens with the $m$ first coupons collected.
As a product space, $\Omega$ is endowed with the product sigma-algebra $\mathcal F=\mathcal K^{\otimes\mathbb N}$, generated by the cylinders $A\times K^\mathbb N$ with $A\subseteq K^n$ and $n\in\mathbb N$, and with the product probability measure $\mu=p^{\otimes\mathbb N}$. Thus, $\mu$ is the unique probability measure such that, for every positive $n$ and every $\eta=(\eta_t)$ in $K^n$, $\mu(\Omega_\eta)=\prod\limits_{t=1}^np_{\eta_t}$, where $\Omega_\eta=\{\omega=(\omega_n)_{n\in\mathbb N}\in\Omega\mid\forall t\leqslant n,\omega_t=\eta_t\}$.
For every $\omega=(\omega_n)_{n\in\mathbb N}$ in $\Omega$ and every $n\in\mathbb N$, define $N_n(\omega)$ as the size of the set $\{\omega_t\mid t\leqslant n\}$. Then each $N_n$ is a random variable on $(\Omega,\mathcal F)$. Furthermore, for every positive $\ell$, the number $T_\ell$ of draws needed to collect exactly $\ell$ different coupons is such that $[T_\ell\leqslant m]=[N_m\geqslant\ell]$ and, for every positive $m$, the number $L_m$ of different coupons collected during the $m$ first draws is $L_m=N_m$. This proves that $[T_\ell\leqslant m]=[L_m\geqslant\ell]$ for every positive $m$ and $\ell$.


TITLE: How do you break up an exact sequence of any length to a "succession of short exact sequences"?
QUESTION [3 upvotes]: Note that if $\text{Hom}_R(D,-)$ functor takes short exact sequences to short exact sequences then it takes exact sequences of any length to exact sequences since any exact sequence can be broken up into a succession of short exact sequences.

Let $A \xrightarrow{\psi} B \xrightarrow{\phi} C$ be exact, then there is the short exact sequence $0 \to \ker \phi \to B \to \text{im} \ \phi \to 0$ but how does that imply $\text{Hom}_R(D,A) \xrightarrow{\psi}' \text{Hom}_R(D,B) \xrightarrow{\phi'} \text{Hom}_R(D,C)$ is exact?

REPLY [3 votes]: A sequence $A\stackrel\psi\to B\stackrel\phi\to C$ is exact iff there exist objects $D$, $E$, $F$, and $G$ and short exact sequences $$0\to D\to A\to E\to 0$$
$$0\to E\to B\to F\to 0$$
$$0\to F\to C\to G\to 0$$
such that the composition $A\to E\to B$ is $\psi$ and the composition $B\to F\to C$ is $\phi$.  It follows that if you apply any functor $T$ which preserves short exact sequences, $T(A)\to T(B)\to T(C)$ is still exact (since you can just apply $T$ to the three short exact sequences above).


TITLE: The concept of continuity in a topological group
QUESTION [0 upvotes]: I am now learning the Lie group theory. People talk about the fundamental group of a topological group. The problem is, how is the continuity defined in a topological group? In other words, in which sense we say two elements are close to each other?
I think we need a metric to define continuity. That is, we first need to define the distance between two arbitrary group elements. But I never see people speaking about the distance between two elements in the $SO(3)$ group.

REPLY [1 votes]: I will give you an answer just for completeness, because Danu pretty much covered it with his comment.
Given two topological spaces $(X,\mathscr{T})$ and $(Y,\mathscr{S})$, we say that a function $f:X\to Y$ is continuous if the pre-image $f^{-1}[A]\in \mathscr{T}$ for any $A\in\mathscr{S}$ ($\mathscr{T},\mathscr{S}$ are the respective collections of open sets).
Also, to define the notion of a limit you just need a topological space. Let $(X,\mathscr{T})$ be a topological space; we call a set $N$ a neighborhood of $x$ if there exists an open set $A\in\mathscr{T}$ with $x\in A\subset N$. Then we say that a sequence (or also a net) $(x_n)_{n\in\mathbb{N}}$ converges to $x\in X$ if for any neighborhood $N$ of $x$, there is a $m\in\mathbb{N}$ such that $x_n\in N$ for any $n\geq m$.
If the topology is $T_2$ or Hausdorff, i.e. separates points, limits can be shown to be unique.
Observe that given a metric/norm/inner product you have automatically a topology "inherited" from these structures, and thus an associated notion of continuity and convergence; nevertheless it is not necessary to define them in order to have these notions.


TITLE: What strategies could be used to prove the validity of this argument in order to not violate restrictions on universal generalization (Hurley)
QUESTION [1 upvotes]: I'm considering a particular argument while working through Hurley's Concise Introduction:
1.  (x)(Ax -> Bx)
2.  (x)(Bx -> (E!y)Cxy)
3.  (x)(Cxy -> Dx)
/ (x)(Ax -> Dx)

I verified the validity of this argument using truth-trees.  However, moving on to actually deriving the conclusion, my current strategy is to use conditional proof:
1.  (x)(Ax -> Bx)
2.  (x)[Bx -> (E!y)Cxy]
3.  (x)(y)(Cxy -> Dx)
4.      | Ax
5.      | Bx
6.      | (E!y)Cxy
7.      | ?
8.      | Dx
9.  Ax -> Dx
10. (x)(Ax -> Dx)

My first thought would be to use existential instantiation:
1.  (x)(Ax -> Bx)
2.  (x)[Bx -> (E!y)Cxy]
3.  (x)(y)(Cxy -> Dx)
4.      | Ax
5.      | Bx
6.      | (E!y)Cxy
7.      | Cxa
8.      | Cxa -> Dx
9.      | Dx
10. Ax -> Dx
11. (x)(Ax -> Dx)

However, this seems to be a violation of the restriction on universal generalization: according to Hurley, "UG must not be used if the instantial variable [x] is free in any preceding line obtained by EI."
By the time the conditional is discharged at line 10, the instance variable x is free in a line resulting from existential instantiation, line 7.
Given that this argument is valid, what other intuitive ways might you approach this problem?  Am I misunderstanding the restriction on universal generalization?
I do want to stress that this is not an assignment.

REPLY [0 votes]: However, this seems to be a violation of the restriction on universal generalization: according to Hurley, "UG must not be used if the instantial variable [x] is free in any preceding line obtained by EI."

Don't be confused by their use of $[x]$.  The important issue is that whatever letter is used, it is the instantiation variable.
Here it is $a$ that is the instantiation variable.  Since this $a$ does not occur free in any line preceding the instantiation, you may use existential generalisation on anything where it occurs consequent of the instantiation.
Further, since $a$ does not even occur in the conclusory $Dx$, existential generalisation is not required.
$${\vdots\\(\forall x)(\forall y)(Cxy\to Dx)\text{ here}\\\quad{\mid\quad{Ax\hspace{34ex}\text{Assumption (arbitrary $x$ occurs free)}\\\vdots\\(\exists!y)Cyx\\(\exists y)(Cxy\wedge(\forall z)(Cxz\to z=y))\hspace{5ex}\text{A Definition of uniqueness}\\\quad{ Cx\color{red}a\wedge(\forall z)(Cxz\to z=\color{red}a)\hspace{9ex}\text{Existential Instance (witnessed by }\color{red}a)\\Cx\color{red}a\hspace{31ex}\text{Simplification}\\Cx\color{red}a\to Dx\hspace{24ex}\text{Universal Instantiation}}\\Dx\hspace{34ex}\textit{Modus Ponens}\text{ ($a$ no longer occurs)}}\\Ax\to Dx}\\(\forall x)(Ax\to Dx)}$$
[It is helpful to use indentation to identify where an instantation variable occurs, to ensure the rules are cleanly followed.  This is optional.]