\section{Introduction (continuous case)}
Consider an elliptic operator
\be L=a(x)\frac{\d^2}{\d x^2}+ b(x)\frac{\d}{\d x} \de
(with $a>0$) on $E:=(-M, N)$\, $(M, N\le \infty)$. Define a function $C(x)$:
$$C(x)=\!\int_{o}^x \frac b a,\qqd x\in E,$$
where $o\in E$ is a reference point. Here and in what follows, the
Lebesgue measure $\d x$ is often omitted. It is convenient for us to
define two measures $\mu$ and $\nu$:
\be \mu(\d x)=\frac{e^{C(x)}}{a(x)}\d x,\qqd
\nu(\d x)= e^{-C(x)}\d x. \de
As usual, the norm on $L^2(\mu)$ is denoted by $\|\cdot\|$. Define
$$\aligned
 {\scr A}(-M, N)&= \text{the
set of absolutely continuous functions
on $(-M, N)$},\\
{\scr A}_0(-M, N)&= \{f\in {\scr A}(-M, N): f\text{ has a compact
support} \},\\
D(f)&=\int_{-M}^N {f'}^2 e^C,\qqd f\in {\scr A}(-M, N),\;\;  M, N\le \infty.
\endaligned
$$
Here $D(f)$ is allowed to be $\infty$.
We are interested in the following eigenvalues:
\begin{align}
\lz^{\text{\rm DD}}&=\inf\{D(f): f\in {\scr A}_0(-M, N), \; \|f\|=1\},\\
\lz^{\text{\rm NN}}&=\inf\{D(f): f\in {\scr A}(-M, N), \;\mu(f)=0,\;
\|f\|=1\},
\end{align}
where  $\mu(f)=\int_E f \d \mu$.

The basic estimates, of $\lz^{\text{\rm DD}}$ for instance, given in \ct{cmf10} are as follows:
\be\big(4\,\kz^{\text{\rm DD}}\big)^{-1}\le \lz^{\text{\rm DD}}\le \big(\kz^{\text{\rm DD}}\big)^{-1},\lb{40}\de
where
\be\big(\kz^{\text{\rm DD}}\big)^{-1}\!\!= \inf_{x<y}
\big[\nu (-M, x)^{-1}\! + \nu (y, N)^{-1}\big]\, \mu (x, y)^{-1},\;\;
\mbox{$\mu(x, y)\!:=\!\!\int_x^y \d\mu$}.\lb{05}\de
The proof for the upper estimate is already straightforward, simply using
the classical variational formula for $\lz^{\text{\rm DD}}$ (cf. \rf{cmf10}{Proof (b) of Theorem 8.2}).
However, the proof for the lower estimate is much harder and deeper, using
capacity theory (cf. \rf{cmf10}{Sections 8, 10}). Even through the capacitary tool is suitable
in a general setup (cf. \ct{fmut}, \rf{cmf05}{Theorems 7.1 and 7.2},
\rf{mav}{Chapter 2}), it is still expected to have a direct proof (avoiding capacity)
in such a concrete situation. This is done at the beginning of the next section.
Surprisingly, the simple proof also works in the ergodic case for which the original proof
is based on (\ref{40}) plus a use of the duality and the coupling technique.
The main body of the paper is devoted to an improvement of the basic lower estimate given in (\ref{40}),
as stated in Corollary \ref{t11} below. The result can be regarded as a typical application  of a recent variational formula
(\rf{cmf11}{Theorem 4.2} or Theorem \ref{t21} below).
This note is an addition to the recent papers \ct{cmf10, cmf11} from which one can find
the motivation of the study on the topic and further references. It is remarkable that the new result makes
the whole analytic proof for the basic estimates more elementary.

Here is our first main result which is a refinement of \rf{cmf11}{Corollary 4.3}.

\crl\lb{t11} {\cms
\begin{itemize}
\item [(1)] We have
$$\lz^{\text{\rm DD}}\ge \big({\underline\kz}^{\text{\rm DD}}\big)^{-1}\ge \big(4\,\kz^{\text{\rm DD}}\big)^{-1},$$
where $\kz^{\text{\rm DD}}$ is given in (\ref{05}) and ${\underline\kz}^{\text{\rm DD}}$ is defined by (\ref{11}) below.
\item [(2)] Let $\mu(-M, N)<\infty$. Then assertion (1) holds if the codes DD are replaced by NN
\big(for instance, $\lz^{\text{\rm NN}}\ge \big({\underline\kz}^{\text{\rm NN}}\big)^{-1}$\big)
and the measures $\mu$ and $\nu$ are exchanged.
\end{itemize}}
\decrl

The remainder of the paper is organized as follows. In the next section,
we present shortly an alternative proof of the estimates in (\ref{40}).
The proof shows one of the main new ideas of the paper. Then we prove Corollary \ref{t11}.
Two illustrating examples are also included in this section.
The discrete analog
of Corollary \ref{t11} is presented in the third section.

\section{Proofs and Examples}

\medskip \noindent {\bf Proof of (\ref{40})}.

Let $\uz\in (-M, N)$ be a reference point. Define
$$\dz_{\uz}^-\!=\!\!\sup_{z\in (-M,\, \uz)}\nu(-M,\, z)\, \mu(z,\, \uz),\qqd
\dz_{\uz}^+\!=\!\!\sup_{z\in (\uz,\, N)}  \mu(\uz,\, z)\,\nu(z,\, N).$$
As will be remarked in the next section, we may assume that
$\dz_{\uz}^{\pm}<\infty$. Otherwise, the problem becomes
either trivial or degenerated. Next, denote by $\lz_{\uz}^{\pm}$ the principal
eigenvalue on $(-M, \uz)$ and $(\uz, N)$, respectively, with common reflecting (Neumann)
boundary at $\uz$ and absorbing (Dirichlet) boundary at $-M$ (and $N$) provided $M< \infty$
($N<\infty$). Actually, by an approximating procedure, one may assume that $M, N<\infty$
(cf. \rf{cmf10}{Proof of Corollary 7.9}).
Next, by a splitting technique, one may choose $\uz=\bar\uz$ to be the unique solution to the
equation $\lz_{\uz}^-=\lz_{\uz}^+$. Then they coincide with $\lz^{\text{\rm DD}}$ since by
\rf{czz03}{Theorem 1.1}, we have
$$\lz_{\uz}^-\wedge \lz_{\uz}^+\le \lz^{\text{\rm DD}}\le \lz_{\uz}^-\vee\lz_{\uz}^+$$
for every $\uz\in (-M, N)$,
where $x\wedge y =\min\{x, y\}$ and dually $x\vee y =\max\{x, y\}$.
Alternatively, $\bar\uz$ is the root of the derivative of the eigenfunction of $\lz^{\text{\rm DD}}$
by \rf{czz03}{Proposition 1.3} and the monotonicity of the eigenfunctions of $\lz_{\bar\uz}^{\pm}$. From now on in this proof, we fix this $\bar\uz$.
For given $\vz>0$, let $\bar x<\bar\uz$ and $\bar y>\bar\uz$ satisfy
$$\nu(-M,\,\bar x)\,\mu(\bar x,\, \bar\uz)\ge \dz_{\bar\uz}^--\vz,\qqd
 \mu(\bar\uz,\, \bar y)\, \nu(\bar y,\, N) \ge \dz_{\bar\uz}^+-\vz,$$
respectively. As a continuous analog of \rf{cmf00}{Theorem 1.1}, we have
$$\big[\big(\lz^{\text{\rm DD}}\big)^{-1}
=\big] \qqd \big(\lz_{\bar\uz}^{+}\big)^{-1}\le 4\,\dz_{\bar\uz}^+\qqd
\big[\le 4  \mu(\bar\uz,\, \bar y)\, \nu(\bar y,\, N)+ 4\vz\big].$$
Hence,
$$\big[\big(\lz^{\text{\rm DD}}\big)^{-1}-4\vz\big]\nu(\bar y,\, N)^{-1}\le 4 \mu(\bar\uz,\, \bar y).$$
In parallel, we have
$$\big[\big(\lz^{\text{\rm DD}}\big)^{-1}-4\vz\big]\nu(-M,\, \bar x)^{-1}\le 4 \mu(\bar x,\,\bar\uz).$$
Summing up the last two inequalities, it follows that
$$\big[\big(\lz^{\text{\rm DD}}\big)^{-1}-4\vz\big]\big[ \nu(-M,\, \bar x)^{-1}
+\nu(\bar y,\, N)^{-1}\big]
\le 4 \mu(\bar x,\,\bar y).$$
That is,
$$\big(\lz^{\text{\rm DD}}\big)^{-1}-4\vz
\le 4 \big[ \nu(-M,\, \bar x)^{-1}
+\nu(\bar y,\, N)^{-1}\big]^{-1}
\mu(\bar x,\,\bar y).$$
In view of (\ref{05}), the right-hand side is bounded from above by $4 {\kz}^{\text{\rm DD}}.$
Since $\vz$ is arbitrary, we have proved the lower estimate in (\ref{40}).
A direct proof for the upper one in (\ref{40}) is presented in \rf{cmf10}{Proof (b) of Theorem 8.2}.
\deprf

\medskip \noindent {\bf Proof of the dual of (\ref{40})}:
$$\big(4\,\kz^{\text{\rm NN}}\big)^{-1}\le \lz^{\text{\rm NN}}\le \big(\kz^{\text{\rm NN}}\big)^{-1},$$
where
$$\big(\kz^{\text{\rm NN}}\big)^{-1}\!\!= \inf_{x<y}
\big[\mu (-M, x)^{-1}\! + \mu (y, N)^{-1}\big]\, \nu (x, y)^{-1}.$$
By exchanging ``Neumann'' and ``Dirichlet'', the splitting
point $\uz=\bar\uz$ is now a common Dirichlet boundary
and $-M$ becomes Neumann boundary if $M<\infty$ (and so is $N$). In other words,
$\bar\uz$ is the unique root of the eigenfunction of $\lz^{\text{\rm NN}}$.
Now, in the proof above, we need only to use \rf{cmf00}{Theorem 3.3}
instead of \rf{czz03}{Theorem 1.1} and making the exchange of $\mu$ and $\nu$.
We have thus returned to the role mentioned in \ct{cmf11}: exchanging the boundary condition
``Neumann'' and ``Dirichlet'' simultaneously
leads to the exchange of the measures $\mu$ and $\nu$.

Here is a direct proof for the upper estimate. Given $x, y\in (-M, N)$ with $x<y$, let $\bar\uz=\bar\uz (x, y)$ be the unique solution to the equation
$$\aligned
&\mu(-M, x)\nu(x, \uz)+\int_x^{\uz}\mu(\d z)\nu(z, \uz)\\
&\qd =\mu(y, N)\nu(\uz, y)+\int_{\uz}^y\mu(\d z)\nu(\uz, z),\qqd \uz\in (x, y).\endaligned$$
Next, define
$$f(z)=-\mathbbold{1}_{\{z\le \bar\uz\}}\nu\big(x\vee z, \bar\uz\big)
+\mathbbold{1}_{\{z>\bar\uz\}}\nu\big(\bar\uz, y\wedge z\big).$$
Then $\mu(f)=0$ by the definition of $\bar\uz$. We have
$$\int_{-M}^N \big|f'\big|^2 e^C
=\nu\big(x, \bar\uz\big)+\nu\big(\bar\uz, y\big)
=\nu\big(x, y\big).$$
Moreover,
$$\aligned
\int_{-M}^N \big(f-\pi(f)\big)^2\d\mu&= \int_{-M}^N f^2\d\mu\\
&> \int_{-M}^x f^2\d\mu+\int_{y}^N f^2\d\mu\\
&=\mu(-M, x)\, \nu\big(x, \bar\uz\big)^2+\mu(y, N)\,\nu\big(\bar\uz, y\big)^2.
\endaligned$$
Note that the function
$$\gz(x)=\az x^2 +\bz (1-x)^2, \qqd x\in (0, 1),\;
\az, \bz>0$$
achieves its minimum $\big(\az^{-1}+\bz^{-1}\big)^{-1}$
at $x^*=(1+\bz/\az)^{-1}$. As an application of this result with
$$\az=\mu(-M, x),\qd \bz=\mu(y, N),\qd x=\nu\big(x, \bar\uz\big)/\nu\big(x, y\big),$$
we get
$$\int_{-M}^N \big(f-\pi(f)\big)^2\d\mu
\ge \frac{\nu(x, y)^2}{\mu(-M, x)^{-1}+\mu(y, N)^{-1}}.$$
Hence
$$\frac{\int_{-M}^N \big(f-\pi(f)\big)^2\d\mu}{\int_{-M}^N \big|f'\big|^2 e^C}
\ge \frac{\nu(x, y)}{\mu(-M, x)^{-1}+\mu(y, N)^{-1}}.$$
Making supremum with respect to $x<y$, we obtain the required
$\kz^{\text{\rm NN}}$.
\deprf

It is remarkable that although the last proof is in parallel to the previous one,
it does not depend on (\ref{40}). This is rather lucky since in other cases,
part (2) of Corollary \ref{t11} for instance, we do not have such a direct
proof.

From now on, unless otherwise stated, we restrict ourselves to the Dirichlet case.
For fixed $\uz$, much knowledge on $\lz_{\uz}^{\pm}$ is known (variational formulas,
approximating procedure and so on, refer to \ct{cmf05, cmf10} for instance).
Of which, only a little is used in the proof above.
For instance, by \rf{czz03}{Corollary 1.5}, we have
$$\Big(\sup_{\uz}\big[\dz_{\uz}^-\wedge \dz_{\uz}^+\big]\Big)^{-1}
\ge \lz^{\text{\rm DD}}
\ge \Big(4\inf_{\uz}\big[\dz_{\uz}^-\vee \dz_{\uz}^+\big]\Big)^{-1}.$$
Thus, if we choose $\bar\uz$ to be the solution of
equation $\dz_{\uz}^-= \dz_{\uz}^+$, then we obtain
$$\big(\dz_{\bar\uz}^-\big)^{-1}
\ge \lz^{\text{\rm DD}}
\ge \big(4\dz_{\bar\uz}^-\big)^{-1}$$
which is even more compact than (\ref{40}) in view of the comparison of $\kz^{\text{\rm DD}}$ and $\dz_{\uz}^{\pm}$.
The problem is that $\bar\uz$, especially the one used in the first proof of this section, is usually not
explicitly known and so a large part of the known results for $\lz_{\bar\uz}^{\pm}$ are not practical.
To overcome this difficulty, the first proof above uses two parameters $x$ and $y$
to get $\kz^{\text{\rm DD}}$ and then to obtain the explicit lower
estimates (\ref{40}). For our main result Corollary {\ref{t11}}, the fixed
point $\bar\uz$ used in the proof of (\ref{40}) is replaced by its mimic given in (\ref{08}) below for suitable
test function $f$. The difference is that equation (\ref{08}) is explicit but
not the one for $\bar\uz$ used in the first proof above.

\medskip \noindent {\bf Proof of Corollary \ref{t11}\,(1)}.

By \ct{cmf10} or \ct{cmf11}, we have known that part (2) of Corollary \ref{t11}
is a dual of part (1). Hence in what follows, we need study part (1) only.

The first inequality in part (1) comes from \rf{cmf11}{Corollary 4.3}.
Thus, it suffices to prove the last inequality in part (1).

Even though it is not completely necessary,
we assume that $M, N<\infty$ until the last paragraph of the proof.

For a given $f\in {\scr C}_+:$
$${\scr C}_+\!=\!\{f\!\in\! {\scr C}(-M, N)\!: f\!>\!0 \;\text{on}\; (-M, N),\; f(-M+0)\!=\!0 \text{ and } f(N-0)\!=\!0\},$$
define
\begin{align}
h^-(z)&\!=\!h_f^-(z)
\!=\!\!\int_{-M}^{z}\! e^{-C(u)}\d u \int_u^{\uz}\!\frac{e^C f}{a},\qqd
 z\le \uz, \\
h^+(z)&\!=\!h_f^+(z)
\!=\!\!\int_{z}^{N}\! e^{-C(u)}\d u \int_{\uz}^u\!\frac{e^C f}{a},\qqd z>\uz, \lb{8}
\end{align}
where $\uz=\uz (f)\in (-M, N)$ is the unique root of the equation:
\be h^-(\uz)= h^+(\uz)\lb{08}\de
provided $h_f^{\pm}<\infty$. The uniqueness of $\uz$ should be clear since on $(-M, N)$,
as a function of $\uz$, $h^-(\uz)$ is continuously increasing from zero to $h^-(N-0)>0$
and $h^+(\uz)$ is continuously decreasing from $h^+(-M+0)>0$ to zero.
Next, define
$$I\!I^{\pm}(f)= h^{\pm}/f.$$
Then we have the following variational formula.
\thm\lb{t21}{\bf\rf{cmf11}{Theorem 4.2\,(1)}}\;\;{\cms  Assume that $\nu(-M, N)<\infty$. Then
\be\lz^{\text{\rm DD}}= \sup_{f\in {\scr C}_+}\Big\{\Big[\inf_{z\in (-m,\,\uz)}I\!I^-(f)(z)^{-1}\Big]
\bigwedge \Big[\inf_{z\in (\uz, N)}I\!I^+(f)(z)^{-1}\Big]\Big\}.\lb{9}\de
}\dethm

We remark that in the original statement of \rf{cmf11}{Theorem 4.2\,(1)}, the boundary
condition ``$f(-M+0)\!=\!0 \text{ and } f(N-0)\!=\!0$'' is ignored. The condition is added
here for the use of the operators $I^{\pm}$ (different from $I\!I^{\pm}$) to be defined later.
However, the conclusion (\ref{9}) remains true since the eigenfunction of $\lz^{\text{\rm DD}}$
does satisfy this condition.

We now fix $x<y$ and let $f=f^{x, y}$:
\be f^{x, y}(s)=
\begin{cases}
\sqrt{\fz^+(y)\fz^-(s\wedge x)/\fz^-(x)},\qd & s\le y\\
\sqrt{\fz^+(s)}, & s\ge y,
\end{cases}
\de
where
$$\fz^-(s)=\nu (-M, s)\qd\text{ and }\qd\fz^+(s)=\nu(s, N).$$
Certainly, here we assume that $\fz^{\pm}<\infty$ (which is automatic whenever $M, N< \infty$).
Clearly, $f^{x, y}\in {\scr C}_+$. Here we are mainly interested in those pair $\{x, y\}$ having the property
$x< \uz< y$.  As proved in \ct{cmf11}, the quantity ${\underline\kz}^{\text{\rm DD}}$:
\begin{align}
{\underline\kz}^{\text{\rm DD}}
&=\inf_{x<y}\Big[\sup_{z\in (-M,\, \uz)}I\!I^-(f^{x, y})(z)\Big]
\bigvee \Big[\sup_{z\in (\uz, N)}I\!I^+(f^{x, y})(z)\Big]\lb{11}
\end{align}
used in Corollary \ref{t11}\,(1) has an explicit expression:
\begin{align}
&\inf_{x< y}
\bigg\{
\sup_{z\in (-M,\, x)}\bigg[\frac{1}{\sqrt{\fz^-(z)}}\,
\mu\Big((\fz^-)^{3/2} \mathbbold{1}_{(-M,\, z)}\Big)
+ \sqrt{{\fz^-(z)}}\,
\mu\Big(\sqrt{\fz^-}\, \mathbbold{1}_{(z,\, x)}\Big)\nonumber\\
&\qqd\qqd\qqd\qd +\sqrt{\fz^-(z)\fz^-(x)}\,\mu(x,\, \uz)\bigg]\nonumber\\
&\qqd\qqd\bigvee \bigg[\frac{1}{\sqrt{\fz^-(x)}}\,
\mu\Big((\fz^-)^{3/2} \mathbbold{1}_{(-M,\, x)}\Big)
+ \mu\Big(\fz^-\, \mathbbold{1}_{(x,\, \uz)}\Big)\bigg]\nonumber\\
&\qqd\qqd \bigvee\sup_{z\in (y,\, N)} \bigg[\frac{1}{\sqrt{\fz^+(z)}}\mu\Big((\fz^+)^{3/2} \mathbbold{1}_{(z,\, N)}\Big)
+ \sqrt{\fz^+(z)}\,\mu\Big(\sqrt{\fz^+}\, \mathbbold{1}_{(y,\, z)}\Big)\nonumber\\
&\qqd\qqd\qqd\qd\qqd +\sqrt{\fz^+(z)\fz^+(y)}\,\mu(\uz,\, y)\bigg]
\bigg\}.\nonumber
\end{align}
We have thus sketched the original attempt (cf. \rf{cmf11}{Corollary 4.3}) to prove Corollary \ref{t11}\,(1). The study was stopped here since we were unable
to compare this long expression with $4\,\kz^{\text{\rm DD}}$.

Before moving further, let us make a remark on (\ref{08}). As proved in \rf{cmf11}{(31)}, for fixed $x$ and $y$, equation (\ref{08}) is
equivalent to the following one.
\begin{align}
\!\!\!\!\frac{\mu\big((\fz^-)^{3/2} \mathbbold{1}_{(-M,\, x)}\big)}{\sqrt{\fz^-(x)}}\,
\!+\! \mu\big(\fz^-\, \mathbbold{1}_{(x,\, \uz)}\big)
\!=\!\frac{\mu\big((\fz^+)^{3/2} \mathbbold{1}_{(y,\, N)}\big)}{\sqrt{\fz^+(y)}}\,
\!+\! \mu\big(\fz^+\, \mathbbold{1}_{(\uz,\, y)}\big)\lb{12}.\end{align}
The quantity in (\ref{12}) is actually the ratio
$$\frac{h^-(\uz)}{f^{x, y}(x)}=\frac{h^+(\uz)}{f^{x, y}(y)}$$
(cf. \rf{cmf11}{(34)}) noting that $f^{x, y}$ is a constant on $[x, y]$:
$$f^{x, y}(x)=f^{x, y}(y)=\sqrt{\fz^+(y)}.$$
Next, note that the left-hand and the right-hand sides of (\ref{12}) are monotone, with respect to $x$ and $y$ respectively, since each of their derivatives does not change its sign:
$$-\frac{e^{-C(x)}}{2(\fz^-)^{3/2}(x)} \mu\big((\fz^-)^{3/2}\mathbbold{1}_{(-M,\, x)}\big)\!\!<\!0\;\text{ and }\;
\frac{e^{-C(y)}}{2 (\fz^+)^{3/2}(y)}\,
\mu\Big((\fz^+)^{3/2} \mathbbold{1}_{(y,\, N)}\Big)\!\!>\!0.$$
The unique solution $\uz$ to (\ref{08}), or equivalently (\ref{12}), should satisfy
\be{\begin{matrix}
\displaystyle\lim_{x\to -M}\frac{\mu\big((\fz^-)^{3/2} \mathbbold{1}_{(-M,\, x)}\big)}{\sqrt{\fz^-(x)}}\,
\!+\! \mu\big(\fz^-\, \mathbbold{1}_{(-M,\, \uz)}\big)
\!\ge\! \frac{\mu\big((\fz^+)^{3/2} \mathbbold{1}_{(\uz,\, N)}\big)}{\sqrt{\fz^+(\uz)}}\,
\qd\text{and}\\
\displaystyle\lim_{y\to N}\frac{\mu\big((\fz^+)^{3/2} \mathbbold{1}_{(y,\, N)}\big)}{\sqrt{\fz^+(y)}}\,
+ \mu\big(\fz^+\, \mathbbold{1}_{(\uz,\, N)}\big)\ge \frac{\mu\big((\fz^-)^{3/2} \mathbbold{1}_{(-M,\, \uz)}\big)}{\sqrt{\fz^-(\uz)}}.
\end{matrix}}\lb{30}\de

As just mentioned above (cf. \rf{cmf11}{(34)}), we also have
\begin{align}
\max_{z\in [x,\,\uz]} I\!I^-\big(f^{x, y}\big)(z)
&= \max_{z\in [\uz, y]} I\!I^+\big(f^{x, y}\big)(z)
=\frac{h^-(\uz)}{f^{x, y}(x)}=\frac{h^+(\uz)}{f^{x, y}(y)}\nonumber\\
&=\frac{1}{\sqrt{\fz^+(y)}}\,
\mu\Big((\fz^+)^{3/2} \mathbbold{1}_{(y,\, N)}\Big)
+ \mu\big(\fz^+\, \mathbbold{1}_{(\uz,\, y)}\big).\lb{13}
\end{align}
Hence we have arrived at
\begin{align}
&\Big[\sup_{z\in (-M,\, \uz)}I\!I^-(f^{x, y})(z)\Big]
\bigvee \Big[\sup_{z\in (\uz, N)}I\!I^+(f^{x, y})(z)\Big]\nonumber\\
&\qd =\Big[\sup_{z\in (-M,\, x)}I\!I^-(f^{x, y})(z)\Big]
\bigvee \frac{h^+(\uz)}{\sqrt{\fz^+(y)}}
\bigvee \Big[\sup_{z\in (y, N)}I\!I^+(f^{x, y})(z)\Big]\lb{14}
\end{align}
which is also known from \ct{cmf11}. Define
$$I^-(f)(x)=\frac{e^{-C(x)}}{f'(x)}\int_x^{\uz} \frac{e^C}{a}f,\qqd
I^+(f)(x)=-\frac{e^{-C(x)}}{f'(x)}\int_{\uz}^x \frac{e^C}{a}f $$
and
$$\dz_{x,\, \uz}^-\!=\!\!\sup_{z\in (-M,\, x)}\fz_z^-\,\mu (z, \uz),\qqd
\dz_{y,\, \uz}^+\!=\!\!\sup_{z\in (y,\, N)}\fz_z^+\,\mu (\uz, z).
$$
Then we have first by the mean value theorem (both $h^-$ and $f^{x, y}$ are
vanished at $-M$) that
$$\sup_{z\in (-M,\, x)}I\!I^-(f^{x, y})(z)\le \sup_{z\in (-M,\, x)}I^-(f^{x, y})(z)$$
and then by \rf{cmf00}{Lemma 1.2} or \rf{cmf05}{page 97} that
$$\sup_{z\in (-M,\, x)}I^-(f^{x, y})(z)\le 4\,\dz_{x,\, \uz}^-.$$
Here we remark that the supremum in the definition of $\dz_{x,\, \uz}^-$
is taken over $(-M,\, x)$ rather than $(-M,\, \uz)\supset (-M,\, x)$.
Hence the original proof for the last estimate needs a slight modification
using the fact that the function $f^{x, y}$ is a constant on $[x,\,\uz]$.
In parallel, since $h^+$ and $f^{x, y}$ vanish at $N$, we have
$$\sup_{z\in (y,\, N)}I\!I^+(f^{x, y})(z)\le \sup_{z\in (y,\, N)} I^+(f^{x, y})(z)
\le 4\,\dz_{y,\, \uz}^+.$$
Therefore, we have arrived at
\begin{align}
{\underline{\kz}^{\rm DD}}
&\le \inf_{x<y}\bigg\{\Big[\sup_{(-M,\, x)}I\!I^-(f^{x, y})\Big]
\bigvee \frac{h^+(\uz)}{\sqrt{\fz^+(y)}}
\bigvee \Big[\sup_{(y,\, N)}I\!I^+(f^{x, y})\Big]\bigg\}\nonumber\\
&\le \inf_{x<\uz<y}\bigg\{\Big[\sup_{(-M,\, x)}I\!I^-(f^{x, y})\Big]
\bigvee \frac{h^+(\uz)}{\sqrt{\fz^+(y)}}
\bigvee \Big[\sup_{(y,\, N)}I\!I^+(f^{x, y})\Big]\bigg\}\nonumber\\
&\le \inf_{x<\uz<y}\bigg\{\big[4\,\dz_{x,\, \uz}^-\big]\bigvee \frac{h^+(\uz)}{\sqrt{\fz^+(y)}}
\bigvee\big[4\,\dz_{y,\, \uz}^+\big] \bigg\}\nonumber\\
&=: \inf_{x<\uz<y} R(x, y,\,\uz)\nonumber\\
&=:\az .\lb{15}
\end{align}
The restriction $\uz\in (x, y)$ is due to the fact that the eigenfunction of ${{\lz}^{\rm DD}}$
is unimodal and $\uz$ is a mimic of its maximum point.
The use of $I\!I^{\pm}$, $I^{\pm}$ and $\dz^{\pm}$ is now
standard (cf. \ct{cmf05}--\ct{cmf11}, for instance).

We now go to the essential new part of the proof. First,
we claim that for each small $\vz$,
there exist $\bar x\in (-M,\,\uz)$ and $\bar y\in (\uz, N)$ (may depend on $\vz$) such that
\be\fz_{\bar x}^-\,\mu ({\bar x}, \uz)\! \ge\! \frac{R(x_0, y_0, \uz_0)}{4}\!-\!\vz,\qqd
\fz_{\bar y}^+\,\mu (\uz, {\bar y})\!\ge\! \frac{R(x_0, y_0, \uz_0)}{4}\!-\!\vz
\lb{16}\de
for some point $(x_0, y_0, \uz_0)$.
In the present continuous case, the conclusion is clear since the infimum
$\az =R(x^*, y^*, \uz^*)$ is achieved at a point $(x^*, y^*, \uz^*)$ with $x^*\le\uz^*\le y^*$, at which
we have not only $h^-(\uz^*)=h^+(\uz^*)$ but also
\be 4\,\dz_{x^*\!,\, \uz^*}^-=4\,\dz_{y^*\!,\, \uz^*}^+= \frac{h^+(\uz^*)}{\sqrt{\fz^+(y^*)}}.\lb{17}\de
To see this, suppose that at the point $(x, y,\,\uz)$ with $x<\uz<y$, we have
\be \frac{h^+(\uz)}{\sqrt{\fz^+(y)}}> 4 \big[\dz_{x,\,\uz}^- \vee \dz_{y,\,\uz}^+\big].\lb{18}\de
Without loss of generality, assume that $\dz_{x,\,\uz}^- \ge \dz_{y,\,\uz}^+$. We now
fix $y$ and let $\tilde \uz \in (\uz, y]$. Then $\dz_{y,\,\uz}^+\ge \dz_{y, \, \tilde\uz}^+$ by definition. In view of (\ref{13}), we have
$$ \frac{h^+(\uz)}{\sqrt{\fz^+(y)}}>\frac{h^+\big(\tilde\uz\big)}{\sqrt{\fz^+(y)}}.$$
Next, to keep $h^-\big(\tilde\uz\big)=h^+\big(\tilde\uz\big)$, one has a new $\tilde x> x$ by using (\ref{12}) (the left-hand side of (\ref{12}) is decreasing in $x$). Correspondingly, we have $\dz_{\tilde x,\, \tilde\uz}^-\ge \dz_{x,\,\uz}^-$.
In particular, for $\tilde \uz$ closed enough to $\uz$ such that
$$\frac{h^+\big(\tilde\uz\big)}{\sqrt{\fz^+(y)}}\ge 4\,\dz_{\tilde x, \, \tilde\uz}^-,$$
we obtain
$$\aligned
\big[4\,\dz_{x,\,\uz}^-\big] \bigvee \frac{h^+(\uz)}{\sqrt{\fz^+(y)}}
\bigvee \big[4\,\dz_{y,\,\uz}^+\big]
&= \frac{h^+(\uz)}{\sqrt{\fz^+(y)}}\\
&>\frac{h^+\big(\tilde\uz\big)}{\sqrt{\fz^+(y)}}\\
&=\big[4\,\dz_{\tilde x,\, \tilde\uz}^-\big] \bigvee  \frac{h^+\big(\tilde\uz\big)}{\sqrt{\fz^+(y)}}
\bigvee \big[4\,\dz_{y,\, \tilde\uz}^+\big].\endaligned$$
Thus, once (\ref{18}) holds, we can find a new point
$(\tilde x, y, \tilde\uz)$ such that $R(x, y, \uz)> R(\tilde x, y, \tilde\uz)$.
In other words, if the infimum $\az $ is attained at $(x^*, y^*, \uz^*)$, we should have
\be \frac{h^+(\uz^*)}{\sqrt{\fz^+(y^*)}}\le 4 \big[\dz_{x^*\!,\, \uz^*}^- \vee \dz_{y^*\!,\, \uz^*}^+\big].\lb{36}\de
One may handle with the other two cases and finally arrive at (\ref{17}).
Note that instead of (\ref{17}), the following weaker condition is still enough for our purpose.
If at some point $(x, y,\,\uz)$,
\be 4\,\dz_{x,\,\uz}^- =4\,\dz_{y,\,\uz}^+=: \az '\ge  \frac{h^+(\uz)}{\sqrt{\fz^+(y)}},\lb{19}
\de
then we have not only $\az '\ge \az $ but also (\ref{16})
for suitable $\bar x\le x$ and $\bar y\ge y$.
To check (\ref{19}), we first mention that the equation $\dz_{x,\,\uz}^- =\,\dz_{y,\,\uz}^+$ is solvable, at least in the case
that $M, N<\infty$. Because $\dz_{x,\, \uz}^-$ starts
from zero at $x=-M$ and then increases as $x\uparrow$;
$\dz_{y,\,\uz}^+$ also starts
from zero at $y=N$ and then increases as $y\downarrow$.
Therefore, there are a lot of $(x, y)$ satisfying the
required equation.
Next, by (\ref{08}), we can regard $\uz$ as a function of $x$ and $y$.
Then, determine $y$ in terms of $x$ by the equation $\dz_{y,\, \uz(x, y)}^+= \dz_{x,\, \uz(x, y)}^-$.
Now there is only one free variable $x$. We claim that (\ref{19}) holds for some $x$ (and then for some $(x, y,\,\uz)$).
Otherwise, the inverse inequality of (\ref{19}) would hold for all $x$ which contradicts with (\ref{36}).

What we actually need is not the pair $\{x, y\}$ satisfying (\ref{19}) but the pair $\{\bar x, \bar y\}$ satisfying (\ref{16}). From which, the remainder of the proof is very much the same as the one given at the beginning of this section. First, we have
$$(\az /4-\vz)\, \fz^- ({\bar x})^{-1}\le \mu ({\bar x}, \uz),\qqd
(\az /4-\vz)\, \fz^+ ({\bar y})^{-1}\le \mu (\uz, {\bar y}).
$$
Summing up these inequalities, we get
$$(\az /4-\vz)\big[\fz^- ({\bar x})^{-1}+\fz^+ ({\bar y})^{-1}\big]
\le \mu ({\bar x}, {\bar y}).$$
Therefore
$$\aligned
\az /4-\vz &\le \big[\fz^- ({\bar x})^{-1}+\fz^+ ({\bar y})^{-1}\big]^{-1}\mu ({\bar x}, {\bar y})\\
&\le \sup_{x<y}\big[\fz^- ({x})^{-1}+\fz^+ ({y})^{-1}\big]^{-1}\mu ({x}, {y})\\
&=\kz^{\text{\rm DD}}\qd(\text{by } (\ref{05})).
\endaligned$$
Combining this fact with (\ref{15}), we obtain
$$
{\underline{\kz}^{\rm DD}}
\le \az   \nonumber
\le 4 \kz^{\text{\rm DD}}+ 4\,\vz.
$$
Letting $\vz\downarrow 0$,
we have thus proved that ${\underline{\kz}}^{\text{\rm DD}}\le 4 \kz^{\text{\rm DD}}$
as required. The main part of the proof is done since the first Dirichlet
eigenvalue is based on compact sets.

Finally, consider the general case that $M, N\le \infty$.
First, we can rule out the degenerated situation that
$\dz_{x, \uz}^-=\dz_{y, \uz}^+=\infty$.
To see this, rewrite $\kz^{\text{\rm DD}}$ as follows
$$\big(\kz^{\text{\rm DD}}\big)^{-1}= \inf_{x<y}
\big[\big(\fz^- (x)\,\mu(x, y)\big)^{-1}  + \big(\mu (x, y)\, \fz^+ (y)\big)^{-1}\big].$$
It is clear that $\big(\kz^{\text{\rm DD}}\big)^{-1}=0$ and then $\lz^{\text{\rm DD}}=0$ by (\ref{40}).
The corollary becomes trivial. Next, if one of $\dz_{x, \uz}^-$ or $\dz_{y, \uz}^+$
is $\infty$, say $\dz_{x, \uz}^-=\infty$ for instance, then
$$\big(\kz^{\text{\rm DD}}\big)^{-1}=  \Big(\sup_y \,\mu (-M, y) \,\fz^+ (y)\Big)^{-1},$$
i.e.,
$$\kz^{\text{\rm DD}}=  \sup_y\, \mu (-M, y)\, \fz^+ (y).$$
This becomes the essentially known one-side Dirichlet problem.
In the case that both of $\dz_{x, \uz}^-$ and $\dz_{y, \uz}^+ $
are finite,
one may adopt an approximating procedure with finite $M$ and $N$.
This was done in the discrete context, refer to
\rf{cmf10}{Proof of Corollary 7.9 and Proof (c) of Theorem 7.10}.
\deprf

\medskip
To illustrate what was going on in the proof above
and the computation/estimation of ${\underline{\kz}^{\rm DD}}$, we
consider two examples to conclude this section.

\xmp{\bf\rf{cmf11}{Example 5.2}}\qd{\rm Consider the simplest example, i.e. the Laplacian operator on $(0, 1)$.
It was proved in \ct{cmf11} that
$\lz^{\text{\rm DD}}=\pi^2$, $\big(\kz^{\text{\rm DD}}\big)^{-1}=16$ and
$\big({\underline\kz}^{\text{\rm DD}}\big)^{-1}\approx 9.43693$.
The eigenfunction of $\lz^{\text{\rm DD}}$ is $g(x)=\sin(\pi x)$ for which
$g'(1/2)=0$ and so $\bar\uz=1/2$ is the root of equation $\lz_{\uz}^-=\lz_{\uz}^+$.
Because of the symmetry, we have $\uz^*=1/2$ and $y^*=1-x^*$.
Since $\mu=\nu=\d x$, we have $\fz_z^-\, \mu(z, 1/2)=(1/2-z) z$. Thus,
$$\dz_x^-=\sup_{z\in (0, x)}\fz_z^-\, \mu(z, 1/2)=\begin{cases}
(1/2-x)x\qd &\text{ if } x\le 1/4\\
1/16& \text{ if } x\in (1/4, 1/2).
\end{cases}$$
By (\ref{13}) and (\ref{12}), we have
$$\frac{h^-(1/2)}{f^{x,\, 1-x}(x)}=\frac 1 8 -\frac{x^2}{10}.$$
Therefore, each
$$x^*\in \bigg[\frac{20-\sqrt{205}}{78},\; \frac{20+\sqrt{205}}{78}\,\bigg]$$
is a solution to the inequality $4\,\dz_x^-\ge {h^-(1/2)}/{f^{x,\, 1-x}(x)}$.
Correspondingly, we have
$$4\,\dz_{x^*}^-=
\begin{cases}
2 (1-2 x^*)x^*  \qd & \text{if } x^*\in \big[\big(20-\sqrt{205}\,\big)/78,\; 1/4\big]\\
1/4 & \text{if } x^*\in \big[1/4,\; \big(20+\sqrt{205}\,\big)/78\big).
\end{cases}$$
Using this, our conclusion that
$${\underline{\kz}}^{\text{\rm DD}}\le 4\,\dz_{x^*}^-\le 4 \kz^{\text{\rm DD}}$$
can be refined as follows:
$$\big(4\,\kz^{\text{\rm DD}}\big)^{-1}=4\le \big(4\,\dz_{x^*}^-\big)^{-1}
\le \frac{35-\sqrt{41/5}}{4}\approx 8.034< 9.43693\approx\big({\underline\kz}^{\text{\rm DD}}\big)^{-1}.$$
It follows that there are many solutions $x^*$, and so we have a lot of freedom in choosing $(\uz^*, x^*, y^*)$ for (\ref{19}).
However, the maximum of $\big(4\,\dz_{x^*}^-\big)^{-1}$ is attained only at the point $x^*$ which is the
smaller root of equation: $4\,\dz_x^-= {h^-(1/2)}/{f^{x,\, 1-x}(x)}$.
}
\dexmp

The next example is unusual since for which the lower bound $\big(4\,\kz^{\text{\rm DD}}\big)^{-1}$
is sharp. Hence, there is no room for the improvement $\big({\underline\kz}^{\text{\rm DD}}\big)^{-1}$.
The proof above seems rather dangerous for this example since at each step
$$\aligned
\big(\lz^{\text{\rm DD}}\big)^{-1}&\le {\underline\kz}^{\text{\rm DD}}
\le \text{Est}(I^{\pm}(f))\le \text{Est}(\dz_{\cdot,\, \uz}^{\pm})\\
&\le 4\big(\ffz (\bar x, \bar y)+\vz\big)
\le 4 \sup_{x<y}\big(\ffz (x, y)+\vz\big)=4\big( {\kz}^{\text{\rm DD}}+\vz\big)
\endaligned$$
for some $\ffz$, where Est\,$(H)$ means the estimate using $H$,
one may lose something. Here we have also explained the reason
why ${\underline{\kz}^{\rm DD}}$ is often much better than
$4\,{\kz}^{\rm DD}$ as shown in the last example.

\xmp{\bf\rf{cmf11}{Example 5.3}}\qd{\rm Consider the operator
$L=\d^2/\d x^2 +b \d/\d x$ with $b>0$ on $(0, \infty)$. It was checked in \ct{cmf11} that
$\lz^{\rm DD}=b^2/4$, $\big(\kz^{\rm DD}\big)^{-1}=b^2$ and
so the lower estimate $\big(4\,\kz^{\rm DD}\big)^{-1}$ is sharp.
The eigenfunction of $\lz^{\text{\rm DD}}$ is $g(x)=x e^{-b x/2}$ for which
$g'(2/b)=0$ and so $\bar\uz=2/b$ solves the equation $\lz_{\uz}^-=\lz_{\uz}^+$.
We have $C(x)= bx$, $\mu (\d x)=e^{b x} \d x$,
$$\fz^-(s)=\int_0^s e^{-b z}\d z=\frac{1}{b}\big(1- e^{-bs}\big)\qd\text{and}\qd
\fz^+(s)=\int_s^\infty e^{-bz}\d z= \frac{1}{b}e^{-bs}.$$
We begin our study on the equation $\dz_{x,\,\uz}^-= \dz_{y,\,\uz}^+$ rather than
Eq.(\ref{12}) since the former one is simpler. Note that the function
$$\fz_z^-\,\mu (z, \uz)=\frac 1 {b^2} \big(1 - e^{-b z}\big) \big(e^{b \uz}-e^{b z}\big),\qqd z\in (0, \uz]$$
achieves its maximum $b^{-2}\big(e^{b\uz/2}-1\big)^2$
at $z=\uz/2$ and the function
$$\mu (\uz, z)\fz_z^+=\frac 1 {b^2} \big(1-e^{b(\uz - z)} \big),\qqd z\ge \uz $$
achieves its maximum $1/b^2$ at $\infty$. Hence
$$\dz_{x,\,\uz}^-= \frac{1}{b^2}\big(e^{b\uz/2}-1\big)^2\qd \forall x\in [\uz/2, \uz]\qd\text{and}\qd
\dz_{y,\,\uz}^+=\frac 1 {b^2}\qd\forall y\ge \uz.$$
Solving the equation
$$\frac{1}{b^2}\big(e^{b\uz/2}-1\big)^2=\frac 1 {b^2},$$
we get
$\uz^*= 2 b^{-1} \log 2.$
To study (\ref{30}), note that
$$\aligned
&\frac{1}{\sqrt{\fz^+(y)}}
\mu\Big((\fz^+)^{3/2} \mathbbold{1}_{(y,\, \infty)}\Big)
+\mu\big(\fz^+ \mathbbold{1}_{(\uz,\, y)}\big)=\frac{2}{b^2}+\frac 1 b (y-\uz)\\
&\frac{1}{\sqrt{\fz^-(x)}}\mu\Big(\!(\fz^-)^{3/2} \mathbbold{1}_{(0,\, x)}\!\Big)
\!+\!\mu\big(\fz^- \mathbbold{1}_{(x,\,\uz)}\big)\\
&\qqd\qqd= \frac{1}{b^2}\bigg\{2-b\uz+e^{b\uz}+bx
-\frac{3\big(b x +\log\big(1+\sqrt{1-e^{-bx}}\,\big)\big)}{2 \sqrt{1-e^{-bx}}}
\bigg\}.
\endaligned$$
Then the second inequality in (\ref{30}) is trivial and the first one there becomes
$$
\frac{1}{b^2}\big(2-b\uz+e^{b \uz}\big)\ge \frac{2}{b^2}.$$
It is now easy to check that $\uz^*=2 b^{-1} \log 2$ does not satisfy this inequality.
In other words, there is no required solution $(x^*, y^*, \uz^*)$ under the restriction $x^*\in [\uz^*/2, \uz^*]$.
Thus, unlike the last example, there is not much freedom in choosing $(x^*, y^*, \uz^*)$ for (\ref{19}).
However, this does not finish the story since the solution $x^*$ may belong to $[0, \uz^*/2)$.

We are now looking for a solution $x^*$ in the interval $[0, \uz^*/2)$.
When $x\le \uz/2$, the maximum of the function
$\sup_{z\le x} \fz_z^-\,\mu (z, \uz)$
on $[0, x]$ is achieved at $x$. Hence
$$\dz_{x,\,\uz}^-= \frac 1 {b^2} \big(1 - e^{-b x}\big) \big(e^{b \uz}-e^{b x}\big)\qd \forall x\in (0, \, \uz/2]\qd\text{and}\qd
\dz_{y,\,\uz}^+=\frac 1 {b^2}\qd\forall y\ge \uz.$$
Solving the equation
$$\frac 1 {b^2} \big(1 - e^{-b x}\big) \big(e^{b \uz}-e^{b x}\big)=\frac 1 {b^2},$$
we obtain
$\uz^*=x-b^{-1} \log\big(1-e^{-b x}\big).$
Besides, solving the equation $4\,\dz_{y^*\!, \, \uz^*}^+=h^+(\uz^*)\big/\!\sqrt{\fz^+(y^*)}\,$,
we get $y^* = 2/b + \uz^*$. Inserting these into Eq.(\ref{12}), we obtain
$$\frac{e^{2 b x}}{e^{b x}-1}+\log \left(1-e^{-b x}\right)=2+
\frac{3}{2 \sqrt{1-e^{-b x}}} \left(b x+2 \log \left(\sqrt{1-e^{-b x}}+1\right)\right).$$
From this, we obtain the required solution $x^*$ as shown by Figures 1 and 2 below,
noting that the constraint that
$x^*\le \uz^*/2$ is equivalent to $x^*\le b^{-1}\log 2$. Having $x^*$ at hand, it is clear that
the solution $\uz^*$ here is very different from $2/b$.
\begin{center}{{{\includegraphics{tmp-1.eps}\hskip-4.8truecm}
}
\vskip-4.2truecm{\hskip-1.8truecm\includegraphics{tmp-2.eps}}\newline
{\bf Figure 1--2}\qd\rm Solution of $x^*=x^*(b)$ when $b$ varies on $(0, 2]$
(the curve on right) and on $[2, 20]$ (the curve on left), respectively.}\end{center}

To see that the solutions $(\bar x, \bar y)$ to (\ref{16}) may not be unique, keeping $\uz^*$ to be the same as in the last paragraph
but replace $y^*$ with a smaller one $\bar y = b^{-1}+\uz^*$, then one can find a point $\bar x$
satisfying Eq.\,(\ref{12}).
}
\dexmp

\section{Birth--death processes (discrete case)}

This section deals with the discrete case which is parallel in principal to the
continuous one studied above, but it is quite involved and so is worth to write down some details here.

The state space is
$$E=\{i\in {\mathbb Z}: -M-1< i< N+1\},\qqd M, N\le \infty.$$ The transition rates
$Q=(q_{ij})$ are as follows: $b_i:=q_{i,i+1}>0$, $a_i:=q_{i, i-1}>0$, $q_{ii}=-(a_i+b_i)$, $i\in E$.
$q_{ij}=0$ for other $i\ne j$.
Thus, we have $a_{-M}>0$ if $M<\infty$ and similarly for $b_N$. The operator of the process becomes
$$\ooz f(i)=b_i\big(f_{i+1}-f_i\big)+ a_i\big(f_{i-1}-f_i\big), \qqd i\in E$$
with a convention $f_{-M-1}=0$ if $M<\infty$ and  $f_{N+1}=0$ if $N<\infty$.
Next, define the speed (or invariant, or symmetric) measure $\mu$ as follows.
Fix a reference point $o\in E$ and set
\begin{gather}
\mu_{o+n}=\frac{a_{o -1} a_{o -2}\cdots a_{o +n+1}}
 {b_{o } b_{o-1 }\cdots b_{o +n}}, \qqd -M-1-o< n\le -2,\nonumber\\
\mu_{o-1 }=\frac{1}{b_{o } b_{o -1}},\qqd
\mu_{o }=\frac{1}{a_{o } b_{o }},\qqd
 \mu_{o+1 }=\frac{1}{a_{o } a_{o +1}},\nonumber\\
 \mu_{o+n }=\frac{b_{o +1} b_{o +2}\cdots b_{o +n-1}}
 {a_{o } a_{o+1 }\cdots a_{o +n}}, \qquad  2\le n<N+1-o.\nonumber
\end{gather}
A change of the reference point $o$ leads to a constant factor only to the sequence
$(\mu_i)$ and so does not make any influence to the results below.
Corresponding to $\ooz$, the Dirichlet form is
\begin{gather}
D(f)=\sum_{-M-1<i\le o} \mu_i a_i (f_i-f_{i-1})^2
        +\sum_{o\le i <N+1} \mu_i b_i (f_{i+1}-f_{i})^2,\nonumber\\
  f\in {\scr K},\; f_{-M-1}=0\;\text{if }M<\infty \text{ and }f_{N+1}=0\;\text{if }N<\infty, \nonumber
\end{gather}
where ${\scr K}$ is the set of functions on $E$ with compact supports. Having these preparations
at hand, one can define the eigenvalues $\lz^{\text{\rm DD}}$ and  $\lz^{\text{\rm NN}}$ on $L^2(\mu)$
as in the first section.

To state our main result in this context, we need more notation. Define
$${\scr C}_+=\{f|_E>0: f_{-M-1}=0\;\text{if }M<\infty \text{ and }f_{N+1}=0\;\text{if }N<\infty\}.$$
Given $f\in {\scr C}_+$, define $h^{\pm}=h_f^{\pm}$ as follows.
$$\aligned
h_i^-&=\sum_{k=-M}^i \frac{1}{\mu_k a_k} \sum_{\ell=k}^{\uz} \mu_{\ell} f_{\ell}
=\sum_{\ell=-M}^{\uz} \mu_{\ell} f_{\ell} \fz_{\ell\wedge i}^-, \qd i\le \uz,\\
h_i^+&=\sum_{k=i}^N \frac{1}{\mu_k b_k} \sum_{\ell=\uz}^{k} \mu_{\ell} f_{\ell}
=\sum_{\ell=\uz}^{N} \mu_{\ell} f_{\ell} \fz_{\ell\vee i}^+, \qd i\ge \uz,
\endaligned$$
where
$$\fz_i^-=\sum_{k=-M}^i \frac{1}{\mu_k a_k},\qqd \fz_k^+=\sum_{\ell=k}^N \frac{1}{\mu_{\ell} b_{\ell}}$$
and $\uz\in (-M-1, N+1)$ will be specified soon.
Applying $h_f^{\pm}$ to the test function $f=f^{m, n}\,(m, n\in E,\, m\le n)$:
$$f_i^{m,n}=
\begin{cases} \sqrt{\fz_n^+ \fz_{i\wedge m}^-/\fz_m^-}\qqd & i\le n\\
   \sqrt{\fz_i^+} & i\ge n,
\end{cases}$$
we obtain a condition for $\uz$ which is an analog of (\ref{30}):
\be{\begin{matrix}
\displaystyle\lim_{m\to -M}\frac{1}{\sqrt{\fz_m^-}}\!
\sum_{k=-M}^{m-1}\! (\fz_k^-)^{3/2}\mu_k \!+\!\! \sum_{k=-M}^{\uz}\! \fz_k^-\mu_k
\displaystyle\!\ge\! \frac{1}{\sqrt{\fz_{\uz}^+}}\!
\sum_{k=\uz}^N  (\fz_k^+)^{3/2}\mu_k\qd \text{and}\\
\displaystyle\lim_{n\to N}\frac{1}{\sqrt{\fz_n^+}}\!
\sum_{k=n+1}^N\!  (\fz_k^+)^{3/2}\mu_k + \sum_{k=\uz}^N  \fz_k^+\mu_k
\displaystyle\!\ge\! \frac{1}{\sqrt{\fz_{\uz}^-}}\!
\sum_{k=-M}^{\uz} (\fz_k^-)^{3/2}\mu_k.
\end{matrix}}\lb{32}\de
However, in the discrete situation, one can not expect (\ref{08}). This leads to a
serious change. To explain the main idea, let us return to Theorem \ref{t21}.
Because the derivative of the eigenfunction of $\lz^{\rm DD}$ has uniquely one zero point,
say $\uz$. We can split the interval $(-M, N)$ into two parts having a common boundary
$\uz$. Thus, the original process is divided into processes having a common reflecting
boundary $\uz$. Theorem \ref{t21} says that the original $\lz^{\rm DD}$ can be represented
by using the principal eigenvalues of these sub-processes. This idea is the starting point of \ct{czz03}, as already used in the first proof in Section 2. Since the maximum point $\uz$ is unknown in advance, in the
original formulation, $\uz$ is free and then there is an additional term $\sup_{\uz}$
in the expression of Theorem \ref{t21}. This term was removed in \ct{cmf11}, choosing
$\uz$ as a mimic of the maximum point of the eigenfunction. Unfortunately, such a mimic still does not work in the
discrete case, we may lost (\ref{08}) and more seriously,
the eigenfunction may be a simple echelon but not a unimodal (cf. \rf{cmf10}{Definition 7.13}).
Therefore, more work is required. Again, the idea goes back to \ct{czz03} except
here the choice of $\uz$ is based on (\ref{32}).
The first key step of the method is constructing two birth--death processes on the left- and
the right-hand sides, separately. As before, the two processes have Dirichlet boundaries
at $-M-1$ and $N+1$ but they now have a common Neumann boundary at $\uz\in E$.
Let us start from the birth--death process with rates $(a_i, b_i)$ and state space $E$.
Fix a constant $\gz>1$.
\begin{itemize}
\item [(L)] The process on the left-hand side has state space $E^{\uz -}=\{i: -M-1<i\le \uz\}$,
reflects at $\uz$ (and so $b_{\uz}=0$).
Its transition structure is the same as the original one except $a_{\uz}$ is replaced by $a_{\uz}^{-, \gz}:=\gz a_{\uz}$.
Then for this process, the sequence $(\mu_i: i\in E^{\uz -})$ is the same as the original one except the original
$\mu_{\uz}$ is replaced by $\mu_{\uz}/\gz$. Hence, the sequence $(\mu_i a_i: i\in E^{\uz -})$
keeps the same as original.
\item [(R)] The process on the right-hand side has state space $E^{\uz +}=\{i: \uz\le i <N+1\}$,
reflects at $\uz$ (and then $a_{\uz}=0$). Its transition structure is again
the same as the original one except $b_{\uz}$ is replaced by $b_{\uz}^{+, \gz}:=\gz (\gz-1)^{-1} b_{\uz}$.
Then for this process, the sequence $(\mu_i: i\in E^{\uz +})$ is the same as the original one except the original
$\mu_{\uz}$ is replaced by $(1-\gz^{-1})\mu_{\uz}$. Hence, the sequence $(\mu_i b_i: i\in E^{\uz +})$
remains the same as original.
\end{itemize}
Noting that $a_{\uz}^{-, \gz} \downarrow a_{\uz}$ and $b_{\uz}^{+, \gz}\uparrow \infty$ as $\gz\downarrow 1$,
$a_{\uz}^{-, \gz} \uparrow \infty$ and $b_{\uz}^{+, \gz}\downarrow b_{\uz}$ as $\gz\uparrow \infty$,
the constant $\gz$ plays a balance role for the principal eigenvalues of these processes.
From here, following the first proof given in Section 2 and using \ct{czz03} and \rf{cmf10}{Theorem 7.10}, one
can prove the basic estimate
$\lz^{\text{\rm DD}}\ge \big(4\kz^{\text{\rm DD}}\big)^{-1}$
in the present context. Certainly, the parallel proof works also in the ergodic case.

We now continue our study on the discrete analog of Corollary \ref{t11}\,(1).
The quantity $\fz^{\pm}$ needs no change. But $h^{\pm}$ has to be modified as follows.
$$\aligned
h_i^{-, \gz}&=\sum_{k=-M}^i \frac{1}{\mu_k a_k} \bigg[\sum_{k\le \ell\le\uz-1} \mu_{\ell} f_{\ell}+\frac{1}{\gz}\mu_{\uz}f_{\uz}\bigg], \qqd i\le \uz,\\
h_i^{+, \gz}&=\sum_{k=i}^N \frac{1}{\mu_k b_k} \bigg[\frac{\gz-1}{\gz}\mu_{\uz}f_{\uz}+
\sum_{\uz+1\le\ell\le k} \mu_{\ell} f_{\ell}\bigg], \qqd i\ge \uz.
\endaligned$$

Finally, define $I\!I^{\pm, \gz}(f)=h^{\pm, \gz}/f$.
It is now more convenient to write the test functions on $E_{\uz}^{\pm}$ separately:
$$
f_i^{-, m}=\sqrt{\fz_{i\wedge m}^-},\qd i\le \uz,\qqd
f_i^{+, n}=\sqrt{\fz_{i\vee n}^+},\qd i\ge \uz.
$$
Comparing with the original $f^{m, n}$, here a factor acting on $f^{-, m}$ is ignored
\big(the reason why one needs the factor in the original case is for $f_{\uz}^{-, m}=f_{\uz}^{+, n}$\big).

\crl\lb{t31} {\cms We have
\be \lz^{\rm DD}\ge \big({\underline\kz}^{\text{\rm DD}}\big)^{-1}\ge \big(4\,\kz^{\text{\rm DD}}\big)^{-1},\lb{20}\de
where
\begin{gather}
\mbox{\hspace{-1em}}{\underline\kz}^{\text{\rm DD}}\!=\inf_{\uz: \text{\rm(\ref{32}) holds}}\,\inf_{m\le \uz\le n}\inf_{\gz>1}\!\Big\{\!\Big[\sup_{ E\owns i\le\uz} I\!I_i^{-, \gz}(f^{-, m})\Big]
\!\bigvee\! \Big[\sup_{\uz\le i\in E} I\!I_i^{+, \gz}(f^{+, n})\Big]\!\Big\},\\
\mbox{\hspace{-2em}}\big(\kz^{\text{\rm DD}}\big)^{-1}\!=\inf_{m, n\in E:\; m \le n}\bigg[\bigg(\sum_{i=-M}^m\frac{1}{\mu_i a_i}\bigg)^{-1}\!
\!\!+ \bigg(\sum_{i=n}^{N}\frac{1}{\mu_i
b_i}\bigg)^{-1}\bigg]\bigg(\sum_{j=m}^n \mu_j\bigg)^{-1}\!\!\!.
\end{gather}
}
\decrl

\prf By using an approximating procedure, one may assume that $M, N<\infty$
(cf. \rf{cmf10}{Proof of Corollary 7.9 and Proof (c) of Theorem 7.10}).
Fix $\uz\in [m, n]$ and define
\begin{gather}
I_i^{-, \gz}(f)=\frac{1}{\mu_i a_i (f_i-f_{i-1})}\bigg[\frac{1}{\gz}\mu_{\uz}+\sum_{i\le \ell\le \uz-1} \mu_{\ell}\bigg],\qqd i\le \uz\nonumber\\
I_i^{+, \gz}(f)=\frac{1}{\mu_i b_i (f_i-f_{i+1})}\bigg[\frac{\gz-1}{\gz}\mu_{\uz}+\sum_{\uz+1\le \ell \le i} \mu_{\ell}\bigg],\qqd i\ge \uz\nonumber\\
\dz_{m,\, \uz}^{-, \gz}=\sup_{i\le m}\fz_i^- \bigg[\frac{1}{\gz}\mu_{\uz}+\!\!\sum_{i\le \ell\le \uz-1} \mu_{\ell}\bigg], \qqd
\dz_{n,\, \uz}^{+, \gz}=\sup_{i\ge n} \fz_i^+\bigg[\frac{\gz-1}{\gz}\mu_{\uz}+\!\!\sum_{\uz+1\le \ell \le i} \mu_{\ell}\bigg].\nonumber
\end{gather}
We have
\begin{align}
\frac{h_{\uz}^{-, \gz}}{f_m^{-, m}}
&=\frac{1}{\sqrt{\fz_m^-}}\sum_{k=-M}^{m-1}  (\fz_k^-)^{3/2}\mu_k
+ \sum_{k=m}^{\uz-1}  \fz_k^- \mu_k +\frac{1}{\gz} \fz_{\uz}^- \mu_{\uz},\\
\frac{h_{\uz}^{+, \gz}}{f_n^{+, n}}
&=\frac{1}{\sqrt{\fz_n^+}}\sum_{k=n+1}^N (\fz_k^+)^{3/2}\mu_k
+ \sum_{k=\uz+1}^{n} \fz_k^+\mu_k +\frac{\gz-1}{\gz} \fz_{\uz}^+ \mu_{\uz}.\end{align}
For simplicity, let
$$H(m, n,\,\uz, \gz)= \max\bigg\{\frac{h_{\uz}^{-, \gz}}{f_m^{-, m}}\mathbbold{1}_{\{m<\uz\}},\;
\frac{h_{\uz}^{+, \gz}}{f_n^{+, n}}\mathbbold{1}_{\{\uz<n\}}\bigg\}.$$
By \rf{cmf10}{Theorem 7.10\,(1), Sections 4, 2 and 3}, we obtain
\begin{align}
\big({{\underline\kz}^{\rm DD}}\big)^{-1}\!\!
&\le\! \inf_{\begin{subarray}{c}\gz>1 \\ [m, n]\owns \uz\\ \uz: \text{\rm(\ref{32}) holds}\end{subarray}}\!\bigg\{\!\bigg[\bigvee_{i\le m}\! I\!I_i^{-, \gz}(f^{-, m})\bigg]
\!\bigvee\! H(m, n,\,\uz, \gz)
\!\bigvee\! \bigg[\bigvee_{i\ge n}\! I\!I_i^{+, \gz}(f^{+, n})\bigg]\!\bigg\}\nonumber\\
&\le \inf_{\begin{subarray}{c}\gz>1 \\ [m, n]\owns \uz\\ \uz: \text{\rm(\ref{32}) holds}\end{subarray}}\bigg\{\!\bigg[\bigvee_{i\le m}\! I_i^{-, \gz}(f^{-, m})\bigg]
\!\bigvee\!H(m, n,\,\uz, \gz)
\!\bigvee\! \bigg[\bigvee_{i\ge n}\! I_i^{+, \gz}(f^{+, n})\bigg]\!\bigg\}\nonumber\\
&\le \inf_{\begin{subarray}{c}\gz>1 \\ [m, n]\owns \uz\\ \uz: \text{\rm(\ref{32}) holds}\end{subarray}}\bigg\{\Big[ 4 \dz_{m, \,\uz}^{-, \gz}\Big]
\bigvee H(m, n,\,\uz, \gz)
\bigvee \Big[4 \dz_{n,\, \uz}^{+, \gz}\Big]\bigg\}\nonumber\\
&=: \inf_{\uz: \text{\rm(\ref{32}) holds}}\,\inf_{[m, n]\owns \uz}\,\inf_{\gz>1}\, R(m, n,\,\uz, \gz)\nonumber\\
&=:\az. \lb{24}
\end{align}
The point we need two terms in the expression of $H$,
rather than one only in the continuous case, is the loss of an analog of (\ref{08}): here we may not have
$h_{\uz}^{-, \gz}=h_{\uz}^{+, \gz}$.
We now choose a candidate of $\uz^*$ (independent of $m$, $n$) from (\ref{32}) and then choose $\{m^*, n^*\}$ with
$m^*\le \uz^*\le n^*$ (may not be unique) so that $(m^*, n^*, \uz^*)$ satisfies the following inequalities
\be{\begin{matrix}\displaystyle
\Big(\dz_{m,\, \uz}^{-, 1}=\!\Big)\;\sup_{i\le m}\fz_i^- \bigg[\mu_{\uz}+\!\!\sum_{i\le \ell\le \uz-1} \mu_{\ell}\bigg]\ge
\sup_{i\ge n} \fz_i^+ \sum_{\uz+1\le \ell \le i} \mu_{\ell}\;\Big(\!= \dz_{n,\, \uz}^{+, 1}\Big)\qd\text{and}\\
\displaystyle\Big(\dz_{m,\, \uz}^{-, \infty}=\!\Big)\;\sup_{i\le m}\fz_i^- \sum_{i\le \ell\le \uz-1} \mu_{\ell} \le
\sup_{i\ge n} \fz_i^+\bigg[\mu_{\uz}+\!\!\sum_{\uz+1\le \ell \le i} \mu_{\ell}\bigg] \;\Big(\!=\dz_{n,\, \uz}^{+, \infty}\Big).\end{matrix}}
\lb{38}\de
Roughly speaking, the condition (\ref{08}) in the continuous case is replaced by a much weaker one (\ref{32}) and the condition $\dz_{m, \,\uz}^{-}= \dz_{n,\, \uz}^{+}$ is replaced by (\ref{38}).
Instead, let $\gz^*$ be the unique solution to the equation
$$\dz_{m^*, \,\uz^*}^{-, \gz}= \dz_{n^*,\, \uz^*}^{+, \gz},\qqd \gz\in [1, \infty]$$
for each fixed pair $\{m^*, n^*\}$: $m^*\le\uz^*\le n^*$.
Here is the balance role played by $\gz$ as mentioned before. As an analog of the continuous case, we are interested in those $\{m^*, n^*\}$: $[m^*\!,\,n^*]\owns \uz^*$ having the property
\be\dz_{m^*\!, \,\uz^*}^{-, \gz^*}= \dz_{n^*\!,\, \uz^*}^{+, \gz^*}\ge\frac{1}{4} \max\bigg\{\frac{h_{\uz^*}^{-, \gz^*}}{f_{m^*}^{-, m^*}}\mathbbold{1}_{\{m^*<\uz^*\}},\;
\frac{h_{\uz^*}^{+, \gz^*}}{f_{n^*}^{+, n^*}}\mathbbold{1}_{\{\uz^*<n^*\}}\bigg\}.\lb{34}\de
Unlike the continuous case, here we may have to repeat the procedure in choosing $(m^*, n^*, \gz^*)$ since $\uz^*$ suggested by (\ref{32}) may not be unique.
Note that the right-hand side of (\ref{34}) is trivial in the particular case that $m^*=n^*=\uz^*$.
Thus, for sufficiently small $\vz>0$, we may choose $(\bar m, \bar n)$ with $[\bar m, \bar n]\owns \uz^*$ and
${\bar\gz}\in (1, \infty)$ such that
\begin{align}
&\fz_{\bar m}^-\bigg[\frac{1}{{\bar\gz}}\mu_{\uz^*}+ \sum_{\ell=\bar m}^{\uz^*-1} \mu_{\ell}\bigg]\ge \frac{R(m^*, n^*, \uz^*, \gz^*)}{4}-\vz \qd\text{and}\nonumber\\
  &\fz_{\bar n}^+\bigg[\frac{{\bar\gz}-1}{{\bar\gz}}\mu_{\uz^*}+\sum_{\ell=\uz^*+1}^{\bar n} \mu_{\ell} \bigg]\ge \frac{R(m^*, n^*, \uz^*, \gz^*)}{4}-\vz. \nonumber\end{align}
Therefore, we have
$$\bigg(\frac{\az}{4}-\vz\bigg)\big(\fz_{\bar m}^-\big)^{-1}\!\!\le
   \frac{1}{{\bar\gz}}\,\mu_{\uz^*}+\! \sum_{\ell=\bar m}^{\uz^*-1} \mu_{\ell},\qqd
\bigg(\frac{\az}{4}-\vz\bigg)\big(\fz_{\bar n}^+\big)^{-1}\!\!\le
   \frac{{\bar\gz}-1}{{\bar\gz}}\,\mu_{\uz^*}+\!\sum_{\ell=\uz^*+1}^{\bar n}\! \mu_{\ell}.
$$
Summing up these inequalities, it follows that
$$\aligned
\bigg(\frac{\az}{4}\!-\!\vz\bigg)
\Big\{\big(\fz_{\bar m}^-\big)^{-1}\!\!+\!
\big(\fz_{\bar n}^+\big)^{-1}\Big\}
\!\le\! \sum_{\ell=\bar m}^{\uz^*-1}\! \mu_{\ell}+\frac{1}{{\bar\gz}}\mu_{\uz^*}\!+\!
\frac{{\bar\gz}-1}{{\bar\gz}}\mu_{\uz^*}\!+\!\!\sum_{\ell=\uz^*+1}^{\bar n}\! \mu_{\ell}
\! =\!\sum_{\ell=\bar m}^{\bar n} \mu_{\ell}.
\endaligned$$
The remainder of the proof is the same as in the continuous situation.\deprf

The following example is almost the simplest one but is indeed very helpful to understand Corollary \ref{t31}
and its proof.

\xmp{\bf \rf{czz03}{Example 2.3} and \rf{cmf10}{Example 7.6\,(2)}}\qd{\rm Let $M=-1$, $N=2$, $b_1=1$, $b_2=2$,
$$a_1=\frac{2-\vz^2}{1+\vz}, \qqd \vz\in \big[0, \sqrt{2}\,\big) \qqd\text{and\qqd $a_2=1$}.$$
Then ${{\lz}^{\rm DD}}=2-\vz$.
It is known that
$$\kz^{\rm DD}=\frac{1}{\lz_0}
-\begin{cases}
{\vz^2}(8 - 4\, \vz^2 + \vz^3)^{-1}\qd &\text{if } \vz\in \big[0,\; \big(\sqrt{13}-1\big)/{3}\big]\\
(8 + 2\, \vz - 3\, \vz^2)^{-1} &\text{if } \vz\in \big[\big(\sqrt{13}-1\big)/{3},\; \sqrt{2}\,\big).
\end{cases}$$

We are now going to compare ${\underline\kz}^{\rm DD}$ with $4\,\kz^{\rm DD}$.
First, we have $\mu_1=\mu_2=1$, $\mu_1 a_1=a_1$, $\mu_1 b_1=b_1$ and $\mu_2 b_2=b_2$.
Next, (\ref{32}) holds for a small part of $\vz$ when $\uz=1$ but never holds if $\uz=2$. Hence, we choose
$\uz=1$. Then $m=1$ and
$$\fz_1^-= \frac{1}{a_1}=\frac{1+\vz}{2-\vz^2},\qqd
\fz_1^+=\frac{1}{b_1}+\frac{1}{b_2}=\frac 3 2, \qqd \fz_2^+=\frac{1}{b_2}=\frac 1 2.$$
Furthermore
$$\dz_{m,\,\uz}^{-, \gz}=\frac{1}{\gz a_1}=\frac{1+\vz}{\gz(2-\vz^2)}.$$
By (\ref{38}), we have $n=1$ or $2$.

(1) When $n=1$, we have
$$\dz_{n,\,\uz}^{+, \gz}
=\bigg[\fz_1^+\bigg(1-\frac{1}{\gz}\bigg)\bigg]\bigvee
\bigg[\fz_2^+\bigg(2-\frac{1}{\gz}\bigg)\bigg]
=\begin{cases}
\displaystyle\frac{3}{2}\bigg(1-\frac{1}{\gz}\bigg)\qd &\text{if } \gz\ge 2\\
\displaystyle 1-\frac{1}{2\gz} &\text{if } \gz\in (1, 2).
\end{cases}$$
Clearly, the equation $\dz_{m,\,\uz}^{-, \gz}=\dz_{n,\,\uz}^{+, \gz}$ has a unique solution
$$\gz=\begin{cases}
\displaystyle\frac{8 + 2 \vz - 3 \vz^2}{3 (2 - \vz^2)}\ge 2 \qd &\text{if $\displaystyle \vz\in\bigg[\frac{\sqrt{13}-1}{3},\, \sqrt{2}\,\bigg)$}\\
\displaystyle\frac{4 + 2 \vz - \vz^2}{2 (2 - \vz^2)}\in (1, 2) \qd &\text{if $\displaystyle \vz\in\bigg(0,\, \frac{\sqrt{13}-1}{3}\bigg)$}.
\end{cases}$$
Correspondingly, with $m=n=\uz=1$, we have
$$\dz_{m,\,\uz}^{-, \gz}=\dz_{n,\,\uz}^{+, \gz}=
\begin{cases}
\displaystyle\frac{3(1+\vz)}{8 + 2 \vz - 3 \vz^2}\qd &\text{if $\displaystyle \vz\in\bigg[\frac{\sqrt{13}-1}{3},\, \sqrt{2}\,\bigg)$}\\
\displaystyle\frac{2(1+\vz)}{4 + 2 \vz - \vz^2}\qd &\text{if $\displaystyle \vz\in\bigg(0,\, \frac{\sqrt{13}-1}{3}\bigg)$}.
\end{cases}$$
It is interesting that the last quantity coincides with $4\,\kz^{\rm DD}$. We have thus arrived at (\ref{34}) since we are in the particular case: $m^*=n^*=\uz^*$.

(2) When $n=2$, we have
$$
\frac{h_{\uz}^{+, \gz}}{f_n^{+, n}}
=\fz_2^+ +\frac{\gz-1}{\gz} \fz_1^+=2-\frac{3}{2\gz},\qqd
\dz_{n,\,\uz}^{+, \gz}
=1-\frac{1}{2\gz}.$$
Clearly,
$$\frac{h_{\uz}^{+, \gz}}{f_n^{+, n}}\le 4\,\dz_{n,\,\uz}^{+, \gz}\qd\text{iff $\gz\ge 1/4$}.$$
As we have seen above, the solution to the equation $\dz_{m,\,\uz}^{-, \gz}=\dz_{n,\,\uz}^{+, \gz}$ is
$$\gz=\frac{4 + 2 \vz - \vz^2}{2 (2 - \vz^2)}>1 \qd \text{on $\big(0, \sqrt{2}\,\big)$}.$$
Then
$$\dz_{m,\,\uz}^{-, \gz}=\dz_{n,\,\uz}^{+, \gz}=\frac{2(1+\vz)}{4 + 2 \vz - \vz^2}>\frac{h_{\uz}^{+, \gz}}{4 f_n^{+, n}}\qd
(m=\uz=1,\; n=2).$$
Hence (\ref{34}) holds.
Combining this case with the last one (i.e., $n=1$), it follows that
${\underline\kz}^{\rm DD}<4\,\kz^{\rm DD}$ for $\vz\in \big(\big(\sqrt{13}-1\big)/3,\, \sqrt{2}\,\big) $.
}\dexmp

\xmp{\bf\rf{cmf10}{Examples 7.7\,(5)}}\qd {\rm Let $E=\{1, 2, \cdots\}$, $a_i=1/i$ and $b_i=1$ for all $i\ge 1$.
Then $\lz^{\rm DD}=(3-\sqrt{5}\,)/2\approx 0.38$ and
 $\big(\lz^{\rm DD}\big)^{-1}\approx 2.618$.
We have
 $\mu_i=i!$, $\mu_i a_i=(i-1)!$ and $\mu_i b_i=i!$
 for all $i\ge 1$. Furthermore, we have
$$\big(\kz^{\rm DD}\big)^{-1}\!\!=\!\!\bigg(\bigg[\sum_{k=1}^1 \frac 1 {(k\! -\! 1)!}\bigg]^{-1}\!\! + \bigg[\sum_{k=4}^{\infty} \frac 1 {k!}
\bigg]^{-1}\bigg)\! \bigg[\sum_{\ell=1}^4 \ell !\bigg]^{-1}\!\!\!
=\!\frac 1 {33}\bigg[1 + \frac{3}{3 e\!-\!8}\bigg]\!\approx 0.6174.$$
And so $\kz^{\rm DD}\approx 1.62$.
With
$$\fz_i^-=\sum_{k=1}^i \frac{1}{\mu_k a_k}=\sum_{k=1}^i \frac{1}{(k-1)!}\qd
\text{and}\qd  \fz_k^+=\sum_{\ell=k}^\infty \frac{1}{\mu_{\ell} b_{\ell}}=\sum_{\ell=k}^\infty \frac{1}{\ell !},$$
we have
\begin{gather}
\dz_{m,\, \uz}^{-, \gz}=\sup_{i\le m}\bigg[\frac{1}{\gz} \uz!+\!\!\sum_{i\le \ell\le \uz-1} \ell!\bigg] \fz_i^-, \qqd
\dz_{n,\, \uz}^{+, \gz}=\sup_{i\ge n} \bigg[\frac{\gz-1}{\gz}\uz!+\!\!\sum_{\uz+1\le \ell \le i} \ell!\bigg] \fz_i^+.\nonumber\\
\frac{h_{\uz}^{-, \gz}}{f_m^{-, m}}
=\frac{1}{\sqrt{\fz_m^{-, \gz}}}\sum_{k=1}^{m-1}  (\fz_k^-)^{3/2} k!
+ \sum_{k=m}^{\uz-1}  \fz_k^- k! +\frac{1}{\gz} \fz_{\uz}^- \uz !,\nonumber\\
\frac{h_{\uz}^{+, \gz}}{f_n^{+, n}}
=\frac{1}{\sqrt{\fz_n^{+, \gz}}}\sum_{k=n+1}^\infty (\fz_k^+)^{3/2} k!
+ \sum_{k=\uz+1}^{n} \fz_k^+ k! +\frac{\gz-1}{\gz} \fz_{\uz}^+ \uz!.\nonumber
\end{gather}
For convenience, let $(L)$, $(R)$, $(M_-)$  and $(M_+)$ denote the last four quantities.

The candidates given by (\ref{32}) are $\uz=2,\,3$.
The case of $\uz=3$ is ruled out by (\ref{38}) and so we fix $\uz=2$. Then with $m=1,\,2$, $(m, n)$ satisfies (\ref{38}) for every $n$: $2\le n \le 17$. For the simplest choice
$m=n=\uz$, $\dz_{m,\, \uz}^{-, \gz}$ is attained at $i=1$
once $\gz\ge 2$, $\dz_{n,\, \uz}^{+, \gz}$ is attained at
$i=4$ whenever $\gz\ge 5/4$, and then the solution to the equation $\dz_{m,\, \uz}^{-, \gz}=\dz_{n,\, \uz}^{+, \gz}$ is $\gz\approx 3.2273$. Therefore,
$$\dz_{m,\, \uz}^{-, \gz}=\dz_{n,\, \uz}^{+, \gz}\approx 1.62.$$
A better choice is $(m, n)=(1, 5)$. Then
$\gz\approx 3.944$ and
$$4\times (R)=4\times (L)\approx 6.042,\qd (M_-)\approx 2.014, \qd (M_+)\approx 5.54.$$
This certainly implies (\ref{34}).}
\dexmp

\xmp{\bf\rf{cmf10}{Examples 7.7\,(8)}}\qd{\rm Let $E=\{1, 2, \cdots\}$,
$a_1=1$, $a_i=(i-1)^2$ for $i\ge 2$ and $b_i=i^2$ for $i\ge 1$.
Then $\lz^{\rm DD}=1/4=\big(4\,\kz^{\rm DD}\big)^{-1}$.
Once again, this example is dangerous.
Clearly,
$\mu_i=1$, $\mu_i a_i= a_i$ and $\mu_i b_i=b_i$ for all $i\ge 1$.
We have
$$\fz_i^-=1+\sum_{k=1}^{i-1} \frac{1}{k^2},\qqd \fz_i^+ =\sum_{\ell=i}^\infty \frac{1}{\ell^2}$$
and
\begin{gather}
\dz_{m,\, \uz}^{-, \gz}=\sup_{i\le m}\bigg[\frac{1}{\gz}+ \uz-i \bigg]\fz_i^-, \qqd
\dz_{n,\, \uz}^{+, \gz}=\sup_{i\ge n} \bigg[\frac{\gz-1}{\gz}+ i-\uz\bigg]\fz_i^+.\nonumber\\
\frac{h_{\uz}^{-, \gz}}{f_m^{-, m}}
=\frac{1}{\sqrt{\fz_m^-}}\sum_{k=1}^{m-1}  (\fz_k^-)^{3/2}
+ \sum_{k=m}^{\uz-1}  \fz_k^-  +\frac{1}{\gz} \fz_{\uz}^- ,\nonumber\\
\frac{h_{\uz}^{+, \gz}}{f_n^{+, n}}
=\frac{1}{\sqrt{\fz_n^+}}\sum_{k=n+1}^\infty (\fz_k^+)^{3/2}
+ \sum_{k=\uz+1}^{n} \fz_k^+ +\frac{\gz-1}{\gz} \fz_{\uz}^+.\nonumber
\end{gather}
As in the last example, we
use $(L)$, $(R)$, $(M_-)$  and $(M_+)$ to denote the last four quantities.
The only candidate by (\ref{32}) is $\uz=2$ which is fixed now. Then (\ref{38}) holds for all $m=1, 2$ and $n\ge 2$. The key for this example is that
$4\times (R)=4$, independent of $\uz$ and $\gz$. With $m=2$, the maximum of $(L)$
is achieved at $i=1$, it tends to 1 as $\gz\to\infty$. Since $m=\uz$, the term
$(M_-)$ is ignored. Besides, we have $(M_+)<4$ for all $n$: $2\le n\le 58$. Therefore,
(\ref{34}) holds.}
\dexmp

To conclude the paper, we make a remark on the generalization of the
results given here.

\rmk {\rm By using a known technique (cf. \rf{cmf05}{Section 6.7}), the variational
formula and its corollaries for the lower estimate of ${\lz}^{\text{\rm DD}}$ can be
extended to a more general setup (Poincar\'e-type inequalities). The upper estimate
is easier and was given in \rf{cmf10}{the remark above Corollary 8.3}.
}
\dermk


\section{Introduction}
Turing machines can be classified according to their numbers of states and symbols.
It is known (see \cite{WN09} for a survey) that there are universal Turing
machines in the following sets (number of states $\times$ number of symbols):
$$2 \times 18,\ 3 \times 9,\ 4 \times 6,\ 5 \times 5,\ 6 \times 4,\ 9 \times 3,\ 18 \times 2.$$
On the other hand, all the Turing machines in the following sets are decidable:
$$1 \times n,\ 2 \times 3,\ 3 \times 2, n \times 1.$$
In order to refine the classification of Turing machines between universal and decidable classes,
properties in connection with the $3x + 1$ function have been considered.

Recall that the $3x + 1$ function $T$ is defined by
$$T(x) =\left\{
\begin{array}{ll}
x/2       & \mbox{ if $x$ is even}\\
(3x +1)/2 & \mbox{ if $x$ is odd}
\end{array}\right.$$
This can also be written $T(2n) = n$, $T(2n + 1) = 3n + 2$.
When function $T$ is iterated on a positive integer, it seems that the loop
$2 \mapsto 1 \mapsto 2$ is always reached, but this is unproven, and is a famous open problem
in mathematics \cite{La10}. For further references, we set
\begin{quote}
{\bf {\boldmath $3 x + 1$} Conjecture}: When function $T$ is iterated from positive integers,
the loop $2 \mapsto 1 \mapsto 2$ is always reached.
\end{quote}
The $3x + 1$ function is also called the Collatz function, and \emph{Collatz-like} functions
are functions on integers with a definition of the following form:
there exist integers $d \ge 2$, $a_i$, $b_i$, $0 \le i \le d-1$, such that, for all integers $x$,
$$f(x) = \frac{a_ix + b_i}{d}\quad \mbox{if}\quad x \equiv i\quad \mbox{(mod $d$)}.$$
With these definitions, we can state the following properties of Turing machines,
that have been used to refine the classification according to the numbers of states and symbols
(see \cite{MM10} for a survey).
\begin{itemize}
\item Turing machines that simulate the iteration of the $3x + 1$ function
and never halt. It is known that there are such machines in the sets
$$2 \times 8,\ 3 \times 5,\ 4 \times 4,\ 5 \times 3,\ 10 \times 2.$$
We improve these results by giving a $3 \times 4$ Turing machine.
\item Turing machines that simulate the iteration of the $3x +1$ function
and halt when the loop $2 \mapsto 1 \mapsto 2$ is reached.
It is known that there is such a machine in the set $6 \times 3$.
In this article, we give four new Turing machines, in the classes
$3 \times 10$, $4 \times 6$, $5 \times 4$ and $13 \times 2$.
\item Turing machine that simulate the iteration of a Collatz-like function.
It is known that there are such machines in the sets
$$2 \times 4,\ 3 \times 3,\ 5 \times 2.$$
\end{itemize}

\section{Preliminaries: Turing machines}
The Turing machines we use have
\begin{itemize}
\item one tape, infinite on both sides, made of cells containing symbols,
\item one reading and writing head,
\item a set $Q = \{A, B, \ldots\}$ of states, plus a halting state $H$ (or $Z$),
\item a set $\Sigma = \{b,0,1,\ldots\}$ of symbols, where $b$ is the blank symbol
(or $\Sigma = \{0,1\}$, when 0 is the blank symbol),
\item a next move function
$$\delta : Q \times \Sigma \rightarrow \Sigma \times \{L, R\} \times(Q \cup\{H\}).$$
\end{itemize}
If $\delta(p,a) = (b,D,q)$, then the Turing machine, reading symbol $a$ in state $p$,
replaces $a$ by $b$, moves in the direction $D \in \{L, R\}$ ($L$ for Left,
$R$ for Right), and comes into state $q$.
On an input $x_k\ldots x_0 \in \Sigma^{k+1}$, the initial configuration is
$^\omega b(Ax_k)\ldots x_0b^\omega$. This means that the word $x_k\ldots x_0$ is written
on the tape between two infinite strings of blank symbols, and the machine is reading
symbol $x_k$ in state $A$.

\begin{table}
$$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|c}
\mbox{symbols} & \multicolumn{13}{c}{}\\                                                \cline{1-3}
10             & Ma    & \mbox{\bf Mi}_2& \multicolumn{11}{c}{}\\                       \cline{1-3}
9              &       &      &\multicolumn{11}{c}{}\\                                  \cline{1-3}
8              & Ba    &      &\multicolumn{11}{c}{}\\                                  \cline{1-3}
7              &       &      &\multicolumn{11}{c}{}\\                                  \cline{1-4}
6              &       & Ma   &  \mbox{\bf Mi}_2 & \multicolumn{10}{c}{}\\          \cline{1-4}
5              &       & Ba   &                  & \multicolumn{10}{c}{}\\          \cline{1-5}
4              &       & Mi_2 & Ma               &\mbox{\bf Mi}_2 & \multicolumn{9}{c}{}\\         \cline{1-6}
3              &       &      &                  & Ma   &\mbox{\bf Mi}_1 & \multicolumn{8}{c}{}\\  \cline{1-13}
2              &       &      &    &      &      &     &     &     & Ba  & Ma & & \mbox{\bf Mi}_2\\  \hline
               & 2     & 3    & 4  & 5    &\,6\, &\,7\,&\,8\,&\,9\,& 10  & 11 & 12 & 13 & \mbox{states}
\end{array}$$
\caption{Turing machines simulating the $3x + 1$ function:
$Ma=$ Margenstern \cite{Ma98,Ma00},
$Ba=$ Baiocchi \cite{Ba98},
$Mi_1=$ Michel \cite{Mi93},
$Mi_2=$ Michel (this paper).
In roman boldface, halting machines.}
\end{table}

\section{The known Turing machines}
Let us give some more precisions about the Turing machines that simulate
the $3x + 1$ function. The following results are displayed in Table 1.

Michel \cite{Mi93} gave a $6 \times 4$ Turing machine that halts when number 1 is reached.
This machine works on numbers written in binary. Division by 2 of even integers
is easy and multiplication by 3 is done by the usual multiplication algorithm.

Margenstern \cite{Ma98,Ma00} gave never halting $5 \times 3$ and $11 \times 2$
Turing machines in binary, and never halting $2 \times 10$, $3 \times 6$,
$4 \times 4$ Turing machines in unary, that is working on numbers $n$ written
as strings of $n$ 1s.

Baiocchi \cite{Ba98} gave five never halting Turing machines in unary, including
$2 \times 8$, $3 \times 5$ and $10 \times 2$ machines that improved
Margenstern's results.

In this article, we give a never halting $3 \times 4$ Turing machine
that works on numbers written in base 3. Multiplication by  3 is easy
and division by 2 is done by the usual division algorithm.
Note that Baiocchi and Margenstern \cite{BM01} already used numbers written in base 3
to define cellular automata that simulate the $3x + 1$ function.

By adding two states to this $3 \times 4$ Turing machine,
we derive a $5 \times 4$ Turing machine that halts when number 1 is reached.

We also give three other Turing machines that halt when number 1 is reached:

\begin{itemize}
\item A $3 \times 10$ Turing machine obtained by adding one state to the
$2 \times 10$ Turing machine of Margenstern \cite{Ma98,Ma00}.
\item A $4 \times 6$ Turing machine obtained by adding one state to the
$3 \times 6$ Turing machine of Margenstern \cite{Ma98,Ma00}.
\item A $13 \times 2$ Turing machine obtained by adding two states to the
$11 \times 2$ Turing machine of Margenstern \cite{Ma98,Ma00}.
\end{itemize}

\section{A never halting $3 \times 4$ Turing machine}
This Turing machine $M_1$ is defined as follows:

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$M_1$ &  $b$  &   0   &   1   &   2 \\
\hline
$A$ & $b$L$C$ & 0R$A$ & 0R$B$ & 1R$A$ \\
\hline
$B$ &  2L$C$  & 1R$B$ & 2R$A$ & 2R$B$ \\
\hline
$C$ & $b$R$A$ & 0L$C$ & 1L$C$ & 2L$C$ \\
\hline
\end{tabular}
\end{center}

The idea is simple. A positive integer is written on the tape, in base 3, in the usual order.
Initially, in state $A$, the head reads the most significant digit, at the left end
of the number. The initial configuration on input $x = \sum_{i=0}^k x_i3^i$ is
$^\omega b(Ax_k)\ldots x_0b^\omega$. Then the machine performs the division by 2, using the usual
division algorithm. Partial quotients are written on the tape. Partial remainders
are stored in the states: 0 in state $A$, 1 in state $B$. When the head passes the right
end of the number, reading a $b$, then
\begin{itemize}
\item if the remainder is 0, nothing is done: $2n \mapsto n$,
\item if the remainder is 1, a 2 is concatenated to the number:
$2n + 1 \mapsto n \mapsto 3n + 2$.
\end{itemize}
Then the head comes back, in state $C$, to the left end of the number and is
ready to perform a new division by 2.

We have the following theorem.
\begin{thm}
The $3x + 1$ conjecture is true iff, for all positive integer
$x = x_k\ldots x_0$ written in base 3, there exists an integer
$n \ge 0$ such that, on input $x_k\ldots x_0$, the Turing machine
$M_1$ eventually reaches the configuration ${^\omega}b0^n(A1)b^\omega$.
\end{thm}

\section{Turing machines that halts on the final loop}
\subsection{A $3 \times 10$ Turing machine}
Margenstern \cite[Fig.\ 11]{Ma00} gave the folowing never halting $2 \times 10$
Turing machine $M_2$.

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$M_2$ &  $b$  &    1    &   $x$   &   $r$   &   $u$
    &  $v$  &    $y$    &   $z$   &   $t$   &   $k$   \\
\hline
$A$ & $b$R$A$ & $x$R$B$ &  1L$A$  & $k$R$B$ & $x$R$A$
& $x$R$A$ & $r$L$A$ & $r$L$A$ & $y$R$A$ &         \\
\hline
$B$ & $z$L$B$ & $u$R$B$ & $x$R$B$ & $y$R$B$ & $v$L$B$
& $u$R$A$ & $t$L$B$ &  1L$A$  & $x$R$B$ & $b$R$B$ \\
\hline
\end{tabular}
\end{center}

Turing machine $M_2$ works on numbers written in unary, so that the initial
configuration on number $n \ge 1$ is $^\omega b(A1)1^{n-1}b^\omega$.
By adding a new state $C$, we can detect the partial configuration
$(A1)b$, and we obtain the following $3 \times 10$ Turing machine $M_3$.

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$M_3$ &  $b$  &    1    &   $x$   &   $r$   &   $u$
    &  $v$  &    $y$    &   $z$   &   $t$   &   $k$   \\
\hline
$A$ & $b$R$A$ & $x$R$C$ &  1L$A$  & $k$R$B$ & $x$R$A$
& $x$R$A$ & $r$L$A$ & $r$L$A$ & $y$R$A$ &         \\
\hline
$B$ & $z$L$B$ & $u$R$B$ & $x$R$B$ & $y$R$B$ & $v$L$B$
& $u$R$A$ & $t$L$B$ &  1L$A$  & $x$R$B$ & $b$R$B$ \\
\hline
$C$ & $b$L$H$ & $u$R$B$ &         & $y$R$B$ &
 &         &         &         &         &         \\
\hline
\end{tabular}
\end{center}

We have the following theorem
\begin{thm}
The $3x + 1$ conjecture is true iff, for all positive integers $n$,
Turing machine $M_3$ halts on the initial configuration $^\omega b(A1)1^{n-1}b^\omega$.
\end{thm}

\subsection{A $4 \times 6$ Turing machine}
Margenstern \cite[Fig.\ 10]{Ma00} gave the folowing never halting $3 \times 6$
Turing machine $M_4$ (Note that transition $(1,z) \mapsto (x\mbox{R}2)$
in this figure should be $(1,z) \mapsto (r\mbox{R}2)$).

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$M_4$ &  $b$  &    1    & $x$   &    $a$  &   $z$   &  $r$ \\
\hline
$A$ & $b$R$A$ & $x$R$B$ & 1L$A$ &  1L$A$  & $r$R$B$ & \\
\hline
$B$ &  1L$B$  & $a$R$C$ & 1L$B$ &  1L$A$  & $x$R$B$ & $b$R$A$ \\
\hline
$C$ & $z$L$C$ & $x$R$C$ & 1L$C$ & $a$R$A$ & $r$R$C$ & $z$L$C$ \\
\hline
\end{tabular}
\end{center}

Turing machine $M_4$ works on numbers written in unary, with initial
configuration ${^\omega}b(A1)1^{n-1}b^\omega$.
By adding a new state $D$, we can detect
the partial configuration $(A1)b$, and we obtain the
following $4 \times 6$ Turing machine $M_5$.

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$M_5$ &  $b$  &    1    & $x$   &    $a$  &   $z$   &  $r$ \\
\hline
$A$ & $b$R$A$ & $x$R$D$ & 1L$A$ &  1L$A$  & $r$R$B$ & \\
\hline
$B$ &  1L$B$  & $a$R$C$ & 1L$B$ &  1L$A$  & $x$R$B$ & $b$R$A$ \\
\hline
$C$ & $z$L$C$ & $x$R$C$ & 1L$C$ & $a$R$A$ & $r$R$C$ & $z$L$C$ \\
\hline
$D$ & $b$L$H$ & $a$R$C$ &       &         & $x$R$B$ &         \\
\hline
\end{tabular}
\end{center}

We have the following theorem.
\begin{thm}
The $3x + 1$ conjecture is true iff, for all positive integers $n$,
Turing machine $M_5$ halts on the initial configuration ${^\omega}b(A1)1^{n-1}b^\omega$.
\end{thm}

\subsection{A $5 \times 4$ Turing machine}
This Turing machine $M_6$ is defined as follows.

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$M_6$ &  $b$  &    0    &    1    &   2 \\
\hline
$A$ & $b$L$C$ &  0R$A$  &  0R$B$  & 1R$A$ \\
\hline
$B$ &  2L$E$  &  1R$B$  &  2R$A$  & 2R$B$ \\
\hline
$C$ & $b$R$D$ &  0L$C$  &  1L$C$  & 2L$C$ \\
\hline
$D$ &         & $b$R$A$ & $b$R$B$ & 1R$A$ \\
\hline
$E$ & $b$R$H$ &  0L$C$  &  1L$C$  & 2L$C$ \\
\hline
\end{tabular}
\end{center}

Turing machine $M_6$ is obtained from Turing machine $M_1$ by adding
a state $D$ that wipes out the useless 0s, and a state $E$ that detects
the partial configuration $b(Bb)$.

We have the following theorem.
\begin{thm}
The $3x + 1$ conjecture is true iff Turing machine $M_6$ halts on all
input $x = x_k\ldots x_0$ representing a positive integer written in base 3.
\end{thm}

\subsection{A $13 \times 2$ Turing machine}
Margenstern \cite[Fig.\ 8]{Ma00} gave the following never halting $11 \times 2$
Turing machine $M_7$ (in this table, $H$ is \emph{not} a halting state).

\begin{center}
\begin{tabular}{|c|c|c|}
\hline
$M_7$ &  0  &   1   \\
\hline
$A$ & 1R$I$ & 0R$B$ \\
\hline
$B$ & 0R$A$ & 0R$G$ \\
\hline
$C$ & 0R$A$ & 1R$D$ \\
\hline
$D$ & 0R$C$ & 1R$E$ \\
\hline
$E$ & 1R$I$ & 1R$F$ \\
\hline
$F$ & 1R$C$ & 0R$G$ \\
\hline
$G$ & 1R$C$ & 1R$H$ \\
\hline
$H$ & 0R$E$ & 1R$G$ \\
\hline
$I$ & 1L$J$ &       \\
\hline
$J$ & 0R$B$ & 1L$K$ \\
\hline
$K$ & 0L$J$ & 1L$J$ \\
\hline
\end{tabular}
\end{center}

This machine works on numbers written in binary, with the least significant bit
at the left end of the number, and digits 0 and 1 coded by 10 and 11,
so that the initial configuration on number $n = x_k\ldots x_0 = \sum_{i=0}^k x_i2^i$
is $^\omega 0(A1)x_01x_1\ldots 1x_k0^\omega$. Division by 2 of even integers
is easy, and multiplication by 3 is done by the usual algorithm.

By adding two new states $L$ and $M$, we can detect the partial configuration
$(A1)10$, and we obtain the following $13 \times 2$ Turing machine $M_8$,
where $Z$ is the halting state.

\begin{center}
\begin{tabular}{|c|c|c|}
\hline
$M_8$ &  0  &   1   \\
\hline
$A$ & 1R$I$ & 0R$L$ \\
\hline
$B$ & 0R$A$ & 0R$G$ \\
\hline
$C$ & 0R$A$ & 1R$D$ \\
\hline
$D$ & 0R$C$ & 1R$E$ \\
\hline
$E$ & 1R$I$ & 1R$F$ \\
\hline
$F$ & 1R$C$ & 0R$G$ \\
\hline
$G$ & 1R$C$ & 1R$H$ \\
\hline
$H$ & 0R$E$ & 1R$G$ \\
\hline
$I$ & 1L$J$ &       \\
\hline
$J$ & 0R$B$ & 1L$K$ \\
\hline
$K$ & 0L$J$ & 1L$J$ \\
\hline
$L$ & 0R$A$ & 0R$M$ \\
\hline
$M$ & 0L$Z$ & 1R$H$ \\
\hline
\end{tabular}
\end{center}

We have the following theorem.
\begin{thm}
The $3x + 1$ conjecture is true iff, for all positive number
$n = x_k\ldots x_0 = \sum_{i=0}^k x_i2^i$, Turing machine $M_8$ halts on the
initial configuration $^\omega 0(A1)x_01x_1\ldots 1x_k0^\omega$.
\end{thm}

\section{Conclusion}
We have given a new $3 \times 4$ never halting Turing machine
that simulates the iteration of the $3x + 1$ function.
It seems that it will be hard to improve the known results
on never halting machines.

On the other hand, for Turing machines that halt on the conjectured
final loop of the $3x + 1$ function, more researches are still to be done.


\section{Introduction}
Discrete approaches have proved useful in many areas of physics.
Regge calculus \cite{Regge} is a discrete formulation of general relativity (GR) where
spacetime is approximated by a triangulated manifold, and the fundamental variables used to describe the metric
are the lengths of the edges of the triangulation. This approach
has been applied with some success to classical gravity \cite{RuthRegge, Gentle}, and used as a starting point for a
lattice quantization of GR \cite{RuthRegge, Rocek, Immirzi, Hamber}.
As other non-perturbative approaches to quantum gravity,
quantum Regge calculus suffers from the problem of defining a unique gauge-invariant measure in the path integral.
The background-independent spinfoam approach \cite{carlo} suggests an original route based on the
well-defined quantum measure of BF theory. The latter is a topological theory where area variables
appear naturally, and whose action can be reduced
to GR by means of so-called simplicity constraints, as discovered by Plebanski \cite{Plebanski}.
This and other motivations have led Rovelli \cite{carloarea} to suggest that 4d quantum gravity should be
related to a modification of Regge calculus where the fundamental variables are the areas
of triangles rather than the edge lengths.
Some effort was put in this line of research by Makela and Williams among others \cite{Makela,Barrett,Wainwright,RuthRegge},
but the problem has been open for more than ten years.
The main difficulty lies in the fact that a generic triangulation has many more triangles than edges,
thus area variables should be constrained. An
explicit expression of these constraints is obscured by their non-local nature in the triangulation.

In this paper, we introduce a description of discrete gravity that overcomes this difficulty.
The key idea is to enlarge the set of variables from areas only, to areas and angles.
In this way the constraints become local, are easy to write explicitly, and further they are related to the
simplicity constraints of Plebanski's formulation of GR.\footnote{Similar ideas have been
investigated by Reisenberger \cite{Mike} and Rovelli \cite{CarloUn}.}

We approximate the spacetime manifold by a simplicial triangulation, where each 4-simplex\footnote{The
4-simplex, also known as pentachoron in the mathematical literature,
is the convex hull of five points. A 4-simplex contains five tetrahedra, ten triangles and ten edges.}
is flat and the curvature is described by deficit angles associated to the triangles.
Regge calculus uses the fact that on each 4-simplex $\sigma$ the ten components of the (constant) metric tensor $g_{\mu\nu}(\sigma)$ can be straighforwardly expressed in terms of the ten edge lengths $\ell_e$.
A further advantage of using the edge lengths as variables is that they
endow each tetrahedron with six quantities which are sufficient
to completely characterize the tetrahedron's geometry. Therefore the gluing of 4-simplices, obtained
by identifying a shared tetrahedron, is trivial and causes no complications.

On a single 4-simplex, there are also ten triangles, suggesting that areas can be
equivalently taken as the metric variables. There are two difficulties with this idea.
First, it is less straighforward to express $g_{\mu\nu}(\sigma)$ in terms of areas.
For instance the change of variables from edge lengths to areas on a 4-simplex
is singular for orthogonal configurations \cite{Barrett}, that is where right angles among the edges
are present.
Even the equal area configuration has a two-fold ambiguity where the same set of areas corresponds to
two different sets of edge lengths.
This is a more significant difficulty than it might seem at a first look, as such configurations are
relevant in the case of a regular lattice, the simplest flat solution to Regge calculus.
The second issue is even more serious.
Ten areas might be enough to describe the 4-geometry of the simplex, but how about
its boundary 3-geometry? Taken any of the five tetrahedra in the 4-simplex, its geometry is
not uniquely defined by the areas of its 4 triangles
(two more quantities are needed, corresponding for instance to (non-opposite) dihedral angles).
So we need the geometry of the full 4-simplex to determine the individual geometry of any of
its boundary tetrahedra. As a consequence, two adjacent 4-simplices in a triangulation will typically induce
different geometries on the common tetrahedron, leading to discontinuities in the metric \cite{Wainwright},
or to non-local constraints involving the two 4-simplices \cite{Makela}.

A solution to the problem can be achieved adding to the areas
enough variables so that the geometry of each of the five tetrahedra can be independently and
completely determined. A natural choice is to add the tetrahedral dihedral angles.
Of course, this pleonastic set of variables needs to be constrained in order
to succesfully reproduce the dynamics of GR. We now turn to the study of these
constraints.

\section{De natura pentachori}
Let us study how to characterize the geometry of a 4-simplex and its five boundary tetrahedra, using
areas and 3d dihedral angles.
We use a notation which might seem counterintuitive at first, but that pays off well in terms of
efficiency and extends to any dimension.
We denote by $V$ the 4-volume of the simplex $\si$ (or the $n$-volume in general), $V(i)$ the 3-volume
of the tetrahedron $\si(i)$ obtained by removing the vertex $i$ from the 4-simplex, $V(ij)$ the area
of the triangle $\si(ij)$ obtained by removing the vertices $i$ and $j$, and so on.
For the dihedral angles, we use the following notation: $\theta_{ij}$ is the 4d dihedral angle
between the tetrahedra $\si(i)$ and $\si(j)$, hinged at the triangle $\si(ij)$;
$\phi_{ij,k}$ is the 3d dihedral angle between the two triangles $\si(ik)$ and $\si(jk)$,
hinged at $\si(ijk)$ within the tetrahedron $\si(k)$; finally,
$\alpha_{ij,kl}$ is the 2d dihedral angle between the edges $\si(ijk)$
and $\si(ijl)$ belonging to the triangle $\si(kl)$.
All dihedral angles are \emph{internal}, thus for instance an equilateral
$4$-simplex has $\cos\theta=1/4$.

These various types of dihedral angles satisfy a number of relations in a closed 4-simplex,
which we present together with their proofs in the Appendix.
An important role in our construction is played by the following expression of the
2d $\alpha$'s in terms of the 3d $\phi$'s,
\begin{eqnarray}\label{a}
\cos\alpha_{ij,kl}=\frac{\cos\phi_{ij,k}+\cos\phi_{il,k} \cos\phi_{jl,k}  }{\sin\phi_{il,k} \sin\phi_{jl,k}}.
\end{eqnarray}
In this formula the 2d angle, belonging to the triangle $kl$, is described in terms of three 3d angles
all belonging to the \emph{same} tetrahedron $k$. In a closed 4-simplex, a triangle is shared by two tetrahedra,
thus there are two possible choices. Consistency of the two choices, i.e. $\alpha_{ij,kl}=\alpha_{ij,lk}$
(see Fig.\ref{fig1}), gives
\begin{eqnarray}\label{aa}
{\cal C}_{kl,ij}(\phi) &\equiv&
\frac{\cos\phi_{ij,k}+\cos\phi_{il,k} \cos\phi_{jl,k}  }{\sin\phi_{il,k} \sin\phi_{jl,k}} \nonumber\\ &-&
\frac{\cos\phi_{ij,l}+\cos\phi_{ik,l} \cos\phi_{jk,l}  }{\sin\phi_{ik,l} \sin\phi_{jk,l}} = 0.
\end{eqnarray}
\begin{figure}[h]
\includegraphics[width=2cm]{aa.eps}
\caption{\label{fig1} The geometric meaning of equation \Ref{aa}: the 2d angle $\alpha_{ij,kl}$ belonging to the shaded
triangle can be expressed in terms of 3d angles associated the thick edges of the tetrahedron $k$, or
equivalenty of the tetrahedron $l$.}
\end{figure}

Thus a consistent gluing of the tetrahedra in a 4-simplex gives relations among the $\phi$'s.
These are three relations per triangle, hence $30$ in total, of which only $20$ are independent.
To see this, we linearized the equations (\ref{aa}) around generic non-degenerate configurations, including the
potentially harmful orthogonal one, and used an algebraic manipulator to study the rank.
The good behaviour of the orthogonal configuration can
also be anticipated by the absence of cosines in the denominator of \Ref{a}.
Of course, our construction would fail for degenerate
configurations where one or more angles equal $0$ or $\pi$.

These relations are important to characterize the geometry of a 4-simplex.
Consider a generic 4-simplex. Its ten 4d angles $\theta_{ij}$ define the Gram matrix
$G_{ij}(\theta) \equiv \cos\theta_{ij}$ (with the convention $\cos\theta_{ii} \equiv -1$).
If the simplex is closed and flat, these ten angles can not be all independent, but have to
satisfy the condition of vanishing of the Gram determinant, $\det G=0$
(e.g. \cite{Barrett1} and \cite{Freidel}).\footnote{This
condition is the origin of the well-known Schl\"afli identity.}
The nine independent quantities parametrize the space of shapes of the 4-simplex
(a scale factor being the tenth and last metric variable).

We then expect that to characterize the geometry in terms of the thirty 3d angles $\phi_{ij,k}$,
there must exist 21 relations among them. These can be found as follows.
First, we consider the Gram matrices $G^k{}_{ij}(\phi)$ associated to the five tetrahedra;
imposing the vanishing of their determinant guarantees that the tetrahedra are closed.
These are five independent conditions.
Next, we use the 2d angle consistency relations \Ref{aa} to ensure
a consistent gluing of the tetrahedra into a 4-simplex.
The complete set
\begin{equation}\label{C1}
\det \,G^k(\phi) = 0, \quad
{\cal C}_{kl,ij}(\phi) = 0
\end{equation}
can be shown, again by linearization, to have rank 21.
Notice that the first constraint is local on each tetrahedron, unlike the second that involves two adjacent tetrahedra.
Hence we found a necessary and sufficient set of relations among
the $\phi$ angles to be the 3d dihedral angles of a 4-simplex.
Other sets are possible (see Appendix B);
the advantage of this one is the transparency of its geometric meaning.

\section{Area-angle Regge calculus}

With the understanding of the geometry of a 4-simplex gained above, we now come to the main point of this paper:
describing the dynamics of general relativity on a discrete manifold, using areas and 3d angles as variables.
For simplicity, we consider here the case of a Riemannian manifold with no boundaries. The extension
to Lorentzian signature and to boundary terms will be discussed elsewhere.

Using the standard notation ($t$ a triangle, $e=tt'$ an edge), the variables
on the full triangulation are $A_t$ and $\phi_e^\tau$.
The gluing conditions \Ref{aa} refer to each pair of edges in a triangle shared by two tetrahedra.
In a triangulation there will be in general many tetrahedra around the same triangle, and \Ref{aa} has to hold
for any choice of two.
However transitivity ensures that it is enough to impose \Ref{aa} to the pairs of tetrahedra belonging to
the same 4-simplex. We can then write these constraints as
\begin{equation}\label{Cee}
{\cal C}^\si_{ee'}(\phi_e^\tau)=0,
\end{equation}
where ${\cal C}^\si_{ee'}$ is
given by \Ref{aa} for $e=\si(kli)$ and $e'=\si(klj)$ sharing a vertex in a 4-simplex $\si$, and zero otherwise.

On a single 4-simplex we have ten areas and thirty 3d angles, thus
we need thirty independent constraints to reduce the total number of variables to ten.
The situation parallels the analysis we performed in the previous section.
We can still take the triangle gluing conditions \Ref{Cee} involving only $\phi$ angles, and
include the areas in the closure conditions for the five tetrahedra.
Denoting $n_t$ the normal to a triangle, we have by definition $|n_t|^2 = A_t^2$ and
$n_t \cdot n_{t'} = - A_t A_{t'} \cos\phi_{tt'}^\tau$. The closure condition on a tetrahedron $\tau$ reads
$N_\tau \equiv \sum_{t\in\tau} n_t = 0$.
By sequentially taking the scalar product of $N_\tau$ with the four $n_t$
we obtain four constraints,
\begin{equation}\label{closure}
{\cal N}_t^\tau(A,\phi) = A_t - \sum_{t'\neq t} A_{t'}\, \cos\phi_{tt'}^\tau = 0.
\end{equation}
Considering the five tetrahedra on the whole 4-simplex \Ref{closure} gives twenty constraints, to
be added to the thirty constraints \Ref{Cee}. Again we studied the number of independent constraints
by linearization, and found that the resulting system has rank 30 for a generic configuration and also for the orthogonal one. Consequently only ten of the 40 variables used are truly independent. This is consistent
with the kinematical degrees of freedom of discrete general relativity.
As shown explicitly in the Appendix, the forty variables $(A_t, \phi_e^\tau)$ satisfying these thirty independent relations determine completely the geometry of the 4-simplex \emph{and} of its five tetrahedra,
thus each tetrahedron has a well-defined geometry, and gluing 4-simplices causes no problems.
In other words, satisfying the constraints \Ref{Cee} and \Ref{closure} allows us to reconstruct uniquely
a set of edge lengths from the variables $(A_t, \phi_e^\tau)$.

We then consider the following action for general relativity,
\begin{eqnarray}\label{action}
&& S[A_{t},\phi_e^\tau,\lambda_t^\tau, \mu^\si_{ee'}] = \sum_{t} A_{t} \, \eps_{t}(\phi) + \\\nonumber
&& \qquad + \sum_\tau \sum_{t\in\tau} \lambda_t^\tau \, {\cal N}_t^\tau(A, \phi) +
\sum_\si \sum_{ee'\in \si} \mu^\si_{ee'} \, {\cal C}^\si_{ee'}(\phi).
\end{eqnarray}
The first term is just the Regge action with independent area-angle variables,\footnote{The
deficit angles $\eps_t$ are given by the sum over 4d angles on the 4-simplices
sharing the triangle $t$, $\eps_t = 2\pi -\sum_{\si} \theta_t^\si$. We describe in the Appendix how
to express them in terms of the $\phi$'s, or in terms of edge lengths as it is done in Regge calculus.}
and the other terms are the constraints \Ref{closure} and \Ref{Cee}
imposed by the Lagrange multipliers $\lambda_t^\tau$ and $\mu_{ee'}^\si$.
As discussed above, they effectively reduce the set of variables $(A_t, \phi_e^\tau)$ to the edge
lenghts $\ell_e$, therefore \Ref{action} is equivalent to the conventional Regge action,
$S_{\rm R}[\ell_e] = \sum_t A_t(\ell_e) \, \eps_t(\ell_e)$.

Notice that our approach should not be seen as a first order formulation of Regge calculus
(see for instance \cite{Caselle, Barrett1}),
because we are adding 3d dihedral angles (the $\phi$'s), not 4d dihedral angles
(the $\theta$'s): only the latter encode the extrinsic curvature of a 3d slice
and are thus conjugate to the areas.

The reader might wonder at this point whether \Ref{action} is a discretization of a continuum action for GR,
like Regge's is a discretization of the Einstein-Hilbert action $\int \sqrt{g} \, R$.
We argue that this is the case, the continuum avatar of \Ref{action} being Plebanski's action \cite{Plebanski}.
The latter is a modified BF action, which schematically reads $S = \int B \wedge F + \mu \, {\cal C}(B)$
(we refer the reader to the literature for more details \cite{Pleb}).
The term ${\cal C}(B)$ is a set of constraints reducing topological BF theory to GR.
We are naturally led towards the interpretation of the first two terms of \Ref{action}
as a discretization of BF theory, with the closure constraint \Ref{closure} implementing
the Gauss constraint of the continuum BF action,
and the third term as the simplicity constraints.
Recall that Plebanski's constraints state that the bi-normal
$B$ to any triangle $\si(ij)$ must be simple, i.e. it must be the wedge product
of (any) two edge vectors. In our notation, $B_{ij} = \pm e_{ijk} \wedge e_{ijl}$ with $k$ and $l$
different from $i$ and $j$. Then if the closure and simplicity constraints are satisfied, they imply
\begin{eqnarray}\label{simpl1}
B_{ij}\cdot B_{ij} &=& e_{ijk}{}^2 \, e_{ijl}{}^2 - (e_{ijk}\cdot e_{ijl})^2 = \nonumber\\ &=&
V_{ijk}{}^2 V_{ijl}{}^2 \sin\alpha_{ijkl}^2 = \nonumber\\ &=& V_{ij}{}^2, \\\label{simpl2}
B_{ij}\cdot B_{ik} &=&
e_{ijk}{}^2 \, (e_{ijl} \cdot e_{ikl}) - (e_{ijk}\cdot e_{ijl}) \, (e_{ijk} \cdot e_{ikl})  \nonumber\\ && \hspace{-1.8cm} =
V_{ijk}{}^2 \, V_{ijl} \, V_{ikl} \, \big( \cos\alpha_{il,jk}-\cos\alpha_{ij,kl} \cos\alpha_{ik,jl} \big) = \nonumber\\ &=&
V_{ij}\, V_{ik} \, \cos\phi_{jk,i},
\end{eqnarray}
thus using \Ref{simpl1} in \Ref{simpl2} gives
\begin{equation}\label{ainv}
\cos\phi_{jk,i} = \frac{\cos\alpha_{il,jk}-\cos\alpha_{ij,kl} \cos\alpha_{ik,jl} }{\sin\alpha_{ij,kl} \sin\alpha_{ik,jl}}.
\end{equation}
This relation can be inverted to give \Ref{a} with \Ref{aa} holding (see the Appendix).
Conversely if \Ref{aa} and \Ref{closure} are satisfied, we can proceed backwards and define
bi-normals satisfying the simplicity constraints.\footnote{Notice that
the simplicity constraints have typically solutions in two sectors. Here we are imposing directly
the solution in the geometric sector, so we do not comment about this ambiguity, which however plays an
important role in quantum models \cite{newvertex, noi, loro}.}

To further study this correspondence, a
canonical analysis of the action \Ref{action} is in progress, and will appear elsewhere \cite{Bianca2}.

\section{Conclusions}
We introduced a modified Regge calculus where the fundamental variables are areas and 3d dihedral angles
between triangles. The action, given in \Ref{action}, is the conventional Regge term
with independent area-angle variables, plus two additional constraints. The first imposes the closure
of each tetrahedron in the triangulation,
the second guarantees a consistent gluing between adjacent tetrahedra, by imposing
the (conformal) geometry of the common triangle to be the same.
All the constraints are local in the triangulation.
We expect our action to be related to a discretization of Plebanski's action.

Our main result is to show that the full set of constraints guarantees
that local variables determine completely the geometry of each
4-simplex \emph{and of each tetrahedron} in the triangulation. As 4-simplices are glued together
identifying a tetrahedron in their boundary, being able to determine the tetrahedra's geometry is
crucial to have a consistent propagation of the degrees of freedom in the triangulation.
The crucial counting of the independent constraints was performed by linearizing the
constraints around generic non-degenerate configurations (see Appendix B). In particular we checked that
the potentially harmful orthogonal configuration is well behaved.
Similarly, the ambiguity of two sets of lengths
giving the same areas \cite{Barrett} is removed in our formalism simply because the two sets give
different 3d angles. On the basis of our analysis, we can not exclude completely the presence
of pathological configurations. However, a well-behaved orthogonal configuration reassures us that
at least for the regular lattice our approach solves both difficulties with area Regge calculus described in the introduction. This opens the way, for instance, to perturbation theory on a flat background.

While studying the constraints we found a number of relations between the dihedral angles of various
dimension of a 4-simplex. We present them together with their derivation in the Appendix.
We also provide an explicit algorithm to compute the edge lengths from area-angles in
a tetrahedron and in a 4-simplex.

We expect our result to have a number of applications, and before concluding,
we would like to briefly point out a few potentially promising ones.

At the classical level, the canonical analysis of \Ref{action} could shed light
on the description of the Hamiltonian algebra of deformation of discrete manifolds \cite{Bianca2}.
The applicability of this approach to numerical studies of lattice gravity has to be
explored. It would for instance be interesting to study whether our approach keeps the good
convergence properties of conventional Regge calculus in the continuum limit \cite{Brewin}.

At the quantum level, there are possible links to the spinfoam formalism that are worth exploring.
The formalism is expected to provide a well-defined measure for a regularized path integral
for non-perturbative quantum gravity (however see also \cite{Bojo}).
Recently a spinfoam model has been proposed \cite{newvertex} (see also \cite{NewImmirzi}),
whose dynamical variables can be expressed as normals to
triangles \cite{noi} (see also \cite{loro}).
The scalar reduction of these quantities produces exactly the variables $(A_t, \phi_e^\tau)$
considered here. The matching of variables suggests that the discrete calculus introduced here is a candidate for
the semiclassical limit of this new spinfoam model, mimicking
what happens in the 3d case with Regge calculus \cite{Ponzano}.
Indeed, the recent advances \cite{Ashtekar} on calculating the graviton propagator from spinfoams \cite{grav}
are based precisely on such a link. In the context of pure area Regge calculus on a single
4-simplex, this idea was investigated in \cite{Bianchi}.

From this viewpoint, it would be useful to study the quantum theory defined on a regular lattice
in perturbative expansion around flat spacetime, as done by Rocek and Williams \cite{Rocek}.

\subsubsection*{Acknowledgements}
We would like to thank Ruth M. Williams for her inspirational work
and Carlo Rovelli for useful discussions.
This research was supported by Perimeter Institute for Theoretical Physics.
Research at Perimeter Institute is supported by the Government of Canada through
Industry Canada and by the Province of Ontario through the Ministry of Research \& Innovation.