1 Introduction
In the study of dynamical systems with an infinite invariant measure, a variety of ergodic and probabilistic limit theorems have been established. They are often related to classical limit theorems for renewal, Markov, or diffusion processes in probability theory. Among this kind of research of dynamical systems, we are going to focus on three distributional limit theorems, the Darling–Kac theorem for occupation times in sets of finite measure, the Dynkin–Lamperti generalized arcsine law for the last time the orbit visits to sets of finite measure, and the Lamperti generalized arcsine law for occupation times in sets of infinite measure, studied by [Reference Aaronson1, Reference Aaronson2, Reference Kocheim and Zweimüller11, Reference Owada and Samorodnitsky14, Reference Sera17, Reference Sera and Yano18, Reference Thaler24, Reference Thaler25, Reference Thaler and Zweimüller27, Reference Zweimüller33]. The aim of the present paper is to establish large deviation estimates related to these limit theorems under similar abstract settings as in [Reference Kocheim and Zweimüller11, Reference Sera and Yano18, Reference Thaler and Zweimüller27, Reference Zweimüller33]. Our abstract results can be applied to, for example, intermittent maps, that is, non-uniformly expanding interval maps with indifferent fixed points. We are motivated by the study of a large deviation estimate related to a generalized arcsine law for occupation times of one-dimensional diffusion processes [Reference Kasahara and Yano10]. We also refer the reader to [Reference Rouault, Yor and Zani16] for another type of large deviations, which is related to the strong arcsine law for a one-dimensional Brownian motion.
In the remainder of this section, we recall known distributional limit theorems and present our large deviation estimates, using Boole’s transformation as a representative example for simplicity. Nonetheless, previous studies as well as our main results are applicable to more general classes of infinite ergodic transformations.
Example 1.1. (Distributional limit theorems for Boole’s transformation)
We refer the reader to [Reference Aaronson3, Reference Adler and Weiss4, Reference Thaler25] for the details of Boole’s transformation. The map
$T:[0,1]\to [0,1]$
given by
$$ \begin{align*} Tx= \begin{cases} x(1-x)/(1-x-x^2), &x\in[0,1/2], \\ 1-T(1-x), &x\in(1/2,1], \end{cases} \end{align*} $$
is conjugated to Boole’s transformation
$\widetilde {T}x=x-x^{-1} (x\in \mathbb {R}\setminus \{0\})$
. Indeed, let
$\phi (x)=(1-x)^{-1}-x^{-1} (x\in (0,1))$
, then
$\widetilde {T}=\phi \circ T \circ \phi ^{-1}$
on
$\mathbb {R}\setminus \{0\}$
. It is easy to see that
$T0=0, T1=1$
,
$T'(0)=T'(1)=1$
,
$T">0$
on
$(0,1/2)$
and
$T"<0$
on
$(1/2,1)$
. In addition, we have
$Tx-x = 1-x -T(1-x)\sim x^3 (x\to 0)$
. Thus, T is a special case of Thaler’s maps, which will be explained in §8. The map T admits the invariant density h given by
$$ \begin{align*} h(x) =\frac{1}{x^2}+\frac{1}{(1-x)^2}, \quad x\in(0,1). \end{align*} $$
Therefore, the invariant measure
$\mu $
given by
$d\mu (x)=h(x)\,dx (x\in [0,1])$
is an infinite measure. Set
$\gamma =\sqrt {2}-1\in (0,1/2)$
, which is a
$2$
-periodic point of T. Indeed,
$T\gamma =1-\gamma \in (1/2,1)$
and hence,
$T^2\gamma =\gamma $
. Let
$$ \begin{align*} A_0=[0,\gamma),\quad Y=[\gamma,T\gamma],\quad A_1=(T\gamma,1]. \end{align*} $$
Then,
$\mu (Y)=\sqrt {2}$
and
$\mu (A_0)=\mu (A_1)=\infty $
. In addition, Y dynamically separates
$A_0$
and
$A_1$
, that is,
$A_i\cap T^{-1}A_j=\emptyset (i\neq j)$
. For a non-negative integer n, a Borel subset
$A\subset [0,1]$
, and
$x\in [0,1]$
, set
$$ \begin{align*} S_n^A(x)=\sum_{k=1}^{n}1_A(T^k x), \quad Z_n^A(x)=\max\{k\leq n:T^k x\in A\}. \end{align*} $$
Here, it is understood that
$\max \emptyset =0$
. In other words,
$S_n^A(x)$
denotes the occupation time in A of the orbit
$\{T^k x\}_{k\geq 0}$
between time
$1$
and n, and
$Z_n^A(x)$
denotes the last time the orbit arrives in A until time n. Fix any Borel probability measure
$\nu (dx)$
absolutely continuous with respect to the Lebesgue measure on
$[0,1]$
. We interpret x as the initial point of the orbit
$\{T^k x\}_{k\geq 0}$
and
$\nu (dx)$
as the initial distribution of the orbit. Then, the Darling–Kac theorem [Reference Aaronson1, Reference Aaronson2] yields that, as
$n\to \infty $
,
$$ \begin{align} \nu\bigg(\frac{\pi S_n^Y}{2\sqrt{n}}\leq t\bigg) \to \frac{2}{\pi}\int_0^t e^{-y^2/\pi}\,dy, \quad t\geq0. \end{align} $$
Next, the Dynkin–Lamperti generalized arcsine law for waiting times [Reference Thaler24] shows that, as
$n\to \infty $
,
$$ \begin{align} \nu\bigg(\frac{Z_n^Y}{n}\leq t\bigg) \to \frac{2}{\pi}\arcsin\sqrt{t}, \quad t\in[0,1]. \end{align} $$
Finally, the Lamperti generalized arcsine law for occupation times [Reference Thaler25] implies that, as
$n\to \infty $
,
$$ \begin{align} \nu\bigg(\frac{S_n^{A_i}}{n}\leq t\bigg) \to \frac{2}{\pi}\arcsin\sqrt{t}, \quad t\in[0,1], \;i=0,1. \end{align} $$
We also remark that convergence rates of (1.1) and (1.2) were also studied in [Reference Melbourne and Terhesiu12, Reference Terhesiu19–Reference Terhesiu21], and a large deviation estimate for the Perron–Frobenius operator related to (1.2) can be found in [Reference Thaler26].
We now illustrate our main results. Our aim is to estimate the left-hand sides of (1.1), (1.2), and (1.3) as
$t\to 0$
.
Example 1.2. (Large deviation estimates for Boole’s transformation)
Under the setting of Example 1.1, we further assume that
$\nu $
is a probability measure supported on
$[\varepsilon , 1-\varepsilon ]$
for some
$\varepsilon \in (0,1/2)$
and admits a Riemann integrable density. Then, there exists some constants
$0<C_1\leq C_2<\infty $
such that, for any positive sequence
$\{c(n)\}_{n\geq 0}$
with
$c(n)\to 0$
and
$c(n)n\to \infty (n\to \infty )$
, the following estimates hold:
$$ \begin{align} C_1 \leq \liminf_{n\to\infty}\frac{p(n)}{\sqrt{c(n)}}\leq \limsup_{n\to\infty}\frac{p(n)}{\sqrt{c(n)}} \leq C_2 \end{align} $$
for
$$ \begin{align*} p(n)= \nu\bigg(\frac{\pi S_n^Y}{2\sqrt{n}}\leq \sqrt{c(n)}\bigg), \,\, \nu\bigg(\frac{Z_n^Y}{n}\leq c(n)\bigg) \quad \text{and} \quad \nu\bigg(\frac{S_n^{A_i}}{n}\leq c(n)\bigg). \end{align*} $$
Note that
$C_1$
and
$C_2$
may depend on
$\nu $
. These estimates are compatible with (1.1), (1.2), and (1.3), respectively, since the right-hand side of (1.1) with
$t=\sqrt {c(n)}$
and those of (1.2) and (1.3) with
$t=c(n)$
are asymptotically equal to
$(2/\pi )\sqrt {c(n)}$
, as
$n\to \infty $
. Nevertheless, (1.1), (1.2), and (1.3) do not imply (1.4) directly.
For the proof, we adopt a method of double Laplace transform as in [Reference Sera and Yano18], imitating the study of generalized arcsine laws for occupation times of one-dimensional diffusion processes [Reference Barlow, Pitman and Yor5, Reference Kasahara and Yano10, Reference Rouault, Yor and Zani16, Reference Watanabe28, Reference Yano29]. Although moment methods were used in [Reference Kocheim and Zweimüller11, Reference Thaler24, Reference Thaler25, Reference Thaler and Zweimüller27, Reference Zweimüller33], double Laplace transform is more adequate for our large deviation estimates. For example, the probability
$\nu (Z_n^Y/n\leq c(n))$
in Example 1.2 has a negligibly small contribution to the kth moment
$\int _{[0,1]} (Z_n^Y /(c(n)n))^k \,d\nu (k=1,2,\ldots )$
, while it has large contributions to the Laplace transform
$$ \begin{align*} \int_{[0,1]} \exp\bigg(-\frac{\unicode{x3bb} Z_n^Y}{c(n)n}\bigg)\,d\nu \quad(\unicode{x3bb}>0) \end{align*} $$
and the double Laplace transform
$$ \begin{align*} \int_0^\infty e^{-qu} \bigg( \int_{[0,1]} \exp\bigg(-\frac{\unicode{x3bb} Z_{[un]}^Y}{c(n)n}\bigg)\,d\nu\bigg)\,du\quad (q,\unicode{x3bb}>0). \end{align*} $$
This is why we adopt a method of double Laplace transform rather than moment methods to estimate
$\nu (Z_n^Y/n\leq c(n))$
.
This paper is organized as follows. In §2, we recall some basic notions of infinite ergodic theory and the theory of regular variation. In §3, we formulate large deviation estimates related to the Darling–Kac theorem and generalized arcsine laws in abstract settings. Section 4 is devoted to introduce some lemmas needed to calculate double Laplace transform. In §§5, 6, and 7, we prove the large deviation estimates by using double Laplace transform. In §8, we apply our abstract results to Thaler’s maps.
2 Preliminaries
Before presenting our main results, let us recall basic concepts of infinite ergodic theory. We basically follow the settings of [Reference Kocheim and Zweimüller11, Reference Sera and Yano18, Reference Thaler and Zweimüller27, Reference Zweimüller33]. We also refer the reader to [Reference Aaronson3] for the foundations of infinite ergodic theory.
Throughout this paper, except in §§1 and 8, we assume the following condition.
-
• Let
$(X,\mathcal {A}, \mu )$
be a
$\sigma $
-finite measure space with
$\mu (X)=\infty $
, and
$T:X \to X$
be a conservative, ergodic, measure-preserving transformation on
$(X,\mathcal {A},\mu )$
, which is abbreviated as CEMPT. In addition, let
$Y\in \mathcal {A}$
with
$\mu (Y)\in (0,\infty )$
.
Let
$\mathbb {N}$
denote the set of all positive integers and set
$\mathbb {N}_0=\mathbb {N}\cup \{0\}$
. For
$A\in \mathcal {A}$
, we write
$1_A$
for the indicator function of A. Since T is a CEMPT, we have
$\sum _{n\geq 0} 1_A\circ T^{n} =\infty $
, almost everywhere (a.e.) for any
$A\in \mathcal {A}$
with
$\mu (A)>0$
. In other words, the orbit
$\{T^n x\}_{n\geq 0}$
visits A infinitely often for
$\mu $
-almost every initial point x. For
$u\in L^1(\mu )$
, define the signed measure
$\mu _u$
on
$(X,\mathcal {A})$
as
$\mu _u(A)=\int _A u\,d\mu (A\in \mathcal {A})$
. The transfer operator
$\widehat {T}:L^1(\mu )\to L^1(\mu )$
is defined by
$\widehat {T}u=d(\mu _u\circ T^{-1})/d\mu (u\in L^1(\mu ))$
. This operator is characterized by the equation
$ \int _X (v\circ T) u \,d\mu = \int _X v(\widehat {T}u)\,d\mu $
for any
$v\in L^{\infty }(\mu )$
and
$u\in L^1 (\mu )$
. The domain of
$\widehat {T}$
can be extended to all non-negative, measurable functions
$u:X\to [0,\infty )$
. Then,
$\int _X \widehat {T}u \, d\mu =\int _X u \, d\mu $
for any non-negative, measurable function u.
We need to extend the concept of uniform sweeping of [Reference Thaler and Zweimüller27, Reference Zweimüller33] slightly. If, for non-negative measurable functions H and G on
$(X,\mathcal {A},\mu )$
, there is some
$C>0$
and
$K\in \mathbb {N}_0$
such that
$C\sum _{k=0}^K \widehat {T}^k H\geq G$
a.e., then H will be called uniformly sweeping (in K steps) for G. Let
$\mathfrak {H}\cup \{G\}$
be a family of measurable functions
$H:X\to [0,\infty )$
. We say
$\mathfrak {H}$
is uniformly sweeping (in K steps) for G if the following condition holds: there exist some constants
$C>0$
and
$K\in \mathbb {N}_0$
such that, for any
$H\in \mathfrak {H}$
, we have
$C\sum _{k=0}^K \widehat {T}^k H\geq G$
a.e.
Let us recall regularly and slowly varying functions. We refer the reader to [Reference Bingham, Goldie and Teugels6] for the details. Let
$f, g: (0,\infty )\to (0,\infty )$
be positive, measurable functions. If
$f(t)/g(t)\to 1 (t\to t_0)$
, then we write
$f(t)\sim g(t)$
(
$t\to t_0$
). We say f is regularly varying of index
$\rho \in \mathbb {R}$
at
$\infty $
(respectively at
$0$
) if, for any
$\unicode{x3bb}>0$
,
$$ \begin{align*} f(\unicode{x3bb} t) \sim \unicode{x3bb}^\rho f(t) \quad(t\to\infty)\text{ (respectively }t\to 0+). \end{align*} $$
In the case where
$\rho =0$
, we say f is slowly varying at
$\infty $
(respectively at
$0$
). A positive sequence
$\{a(n)\}_{n\geq 0}$
is called regularly varying of index
$\rho $
if the function
$a( [t])$
is regularly varying of index
$\rho $
at
$\infty $
. Here,
$[t]$
denotes the greatest integer that is less than or equal to t.
Let
$\varphi :X\to \mathbb {N}\cup \{\infty \}$
be the first return time to Y, that is,
$$ \begin{align*} \varphi(x)=\min\{k\geq1:T^k x\in Y\}\quad(x\in X). \end{align*} $$
Here, it is understood that
$\min \emptyset =\infty $
. Define disjoint sets
$Y_0, Y_1, Y_2,\ldots \in \mathcal {A}$
as
$$ \begin{align*} Y_0=Y,\quad Y_n= (T^{-n}Y)\setminus \bigg(\bigcup_{k=0}^{n-1}T^{-k}Y \bigg)=Y^c\cap\{\varphi=n\} \quad (n\in\mathbb{N}). \end{align*} $$
As proved in [Reference Thaler and Zweimüller27, equation (2.3)],
$$ \begin{align} 1_{Y_n} = \sum_{k>n}\widehat{T}^{k-n}1_{Y\cap \{\varphi=k\}} \quad \text{a.e.\;\;(}n\in\mathbb{N}_0), \end{align} $$
and
$\mu (Y_n)=\mu (Y\cap \{\varphi>n\})$
. Let
$\{w_n^Y\}_{n\geq 0}$
denote the wandering rate of Y, which is given by
$$ \begin{align} w_n^Y &= \mu\bigg(\bigcup_{k=0}^{n-1}T^{-k}Y\bigg) = \sum_{k=0}^{n-1}\mu(Y_k) = \sum_{k=0}^{n-1} \int_Y \widehat{T}^k 1_{Y_k}\,d\mu \nonumber\\ &= \sum_{k=0}^{n-1} \mu(Y\cap\{\varphi>k\}) \quad (n\in{\mathbb{N}}_0). \end{align} $$
Since T is a CEMPT, we see
$\bigcup _{n\geq 0}T^{-n}Y=X$
, a.e. and hence,
$w_n^Y\to \infty $
(
$n\to \infty $
). For
$s>0$
, let
$Q^Y(s)$
be a Laplace transform of
$\{w_{n+1}^Y-w_n^Y\}_{n\geq 0}$
:
$$ \begin{align*} Q^Y(s)=\sum_{n\geq0} e^{-ns}(w_{n+1}^Y-w_n^Y) =\sum_{n\geq0}e^{-ns}\mu(Y\cap\{\varphi>n\}) \quad (s>0). \end{align*} $$
Then,
$0<Q^Y(s)<\infty $
and
$Q^Y(s)\to \infty $
(
$s\to 0+$
). Let
$\alpha \in (0,1)$
and let
$\ell :(0,\infty )\to (0,\infty )$
be a positive, measurable function slowly varying at
$\infty $
. By Karamata’s Tauberian theorem [Reference Bingham, Goldie and Teugels6, Theorem 1.7.1], the condition
$$ \begin{align*} w_n^Y \sim n^{1-\alpha}\ell(n) \quad (n\to\infty) \end{align*} $$
is equivalent to
$$ \begin{align} Q^Y(s)\sim \Gamma(2-\alpha) s^{-1+\alpha} \ell(s^{-1}) \quad (s\to 0+). \end{align} $$
Here,
$\Gamma (z)=\int _0^\infty e^{-t}t^{-1+z}\,dt$
(
$z>0$
) denotes the gamma function.
If
$\{(w_n^Y)^{-1}\sum _{k=0}^{n-1}\widehat {T}^k 1_{Y_k}\}_{n\geq 1}$
converges in
$L^\infty (\mu )$
as
$n\to \infty $
, then we call the limit function
$H\in L^\infty (\mu )$
as the asymptotic entrance density of Y. Since
$(w_n^Y)^{-1}\sum _{k=0}^{n-1}\widehat {T}^k 1_{Y_k}$
is a
$\mu $
-probability density function supported on Y, so is H. Let
$G\in \{u\in L^1(\mu ):u\geq 0\}$
. Then, H is uniformly sweeping in K steps for G if and only if there exists
$N\in \mathbb {N}$
such that
$\{(w_n^Y)^{-1}\sum _{k=0}^{n-1}\widehat {T}^k 1_{Y_k}\}_{n\geq N}$
is uniformly sweeping in K steps for G.
For non-negative sequences
$(a(n))_{n\geq 0}$
and
$(b(n))_{n\geq 0}$
, we write
$a(n)=o(b(n)) (n\to \infty )$
if, for any
$\varepsilon>0$
, there exists
$n_0\in \mathbb {N}_0$
such that
$a(n)\leq \varepsilon b(n)$
for any
$n> n_0$
.
Lemma 2.1. Assume that there exists
$H\in L^\infty (\mu )$
such that
$$ \begin{align*} \|\widehat{T}^n 1_{Y\cap \{\varphi=n\}} - \mu(Y\cap \{\varphi=n\})H \|_{L^\infty(\mu)} = o(\mu(Y\cap \{\varphi=n\})) \quad (n\to\infty). \end{align*} $$
Then, H is the asymptotic entrance density of Y.
Proof. Fix
$\varepsilon>0$
arbitrarily. Take
$n_0\in \mathbb {N}_0$
large enough so that
$$ \begin{align} \|\widehat{T}^n 1_{Y\cap \{\varphi=n\}} - \mu(Y\cap \{\varphi=n\})H \|_{L^\infty(\mu)} \leq \varepsilon\mu(Y\cap \{\varphi=n\}) \quad \text{for any }n>n_0. \end{align} $$
$$ \begin{align} &\bigg\|\sum_{k=0}^{n-1}\widehat{T}^k 1_{Y_k}-w_n^Y H\bigg\|_{L^\infty(\mu)} \nonumber\\ &\quad= \bigg\|\sum_{k=0}^{n-1}\sum_{m=k+1}^\infty (\widehat{T}^m 1_{Y\cap \{\varphi=m\}} - \mu(Y\cap \{\varphi=m\})H) \bigg\|_{L^\infty(\mu)} \nonumber\\ &\quad\leq \sum_{k=0}^{n_0-1}\sum_{m=k+1}^{n_0} \|\widehat{T}^m 1_{Y\cap \{\varphi=m\}} - \mu(Y\cap \{\varphi=m\})H\|_{L^\infty(\mu)} \nonumber\\ &\qquad+ \sum_{k=0}^{n-1}\sum_{m=\max\{k,n_0\}+1}^{\infty} \|\widehat{T}^m 1_{Y\cap \{\varphi=m\}} - \mu(Y\cap \{\varphi=m\})H\|_{L^\infty(\mu)}. \end{align} $$
It follows from (2.4) and (2.2) that
$$ \begin{align} &\sum_{k=0}^{n-1}\sum_{m=\max\{k,n_0\}+1}^{\infty} \|\widehat{T}^m 1_{Y\cap \{\varphi=m\}} - \mu(Y\cap \{\varphi=m\})H\|_{L^\infty(\mu)} \nonumber\\ &\quad\leq \varepsilon \sum_{k=0}^{n-1}\sum_{m=k+1}^{\infty} \mu(Y\cap \{\varphi=m\}) = \varepsilon w_n^Y. \end{align} $$
Since
$w_n^Y\to \infty (n\to \infty )$
, we use (2.5) and (2.6) to obtain
$$ \begin{align*} \limsup_{n\to\infty} \bigg\|\frac{1}{w_n^Y}\sum_{k=0}^{n-1}\widehat{T}^k 1_{Y_k}-H\bigg\|_{L^\infty(\mu)} \leq \varepsilon. \end{align*} $$
The proof is complete, since
$\varepsilon>0$
was arbitrary.
3 Main results
In the following, we are going to formulate three types of large deviation estimates, which are related to already-known distributional limit theorems.
3.1 Large deviation estimates related to the Dynkin–Lamperti generalized arcsine law
Let
$u:X\to [0,\infty )$
be a non-negative,
$\mu $
-integrable function. Recall
$\mu _u$
is defined as the
$\mu $
-absolutely continuous finite measure on X with density function u with respect to
$\mu $
, that is,
$$ \begin{align*} \mu_{u}(A) =\int_A u(x)\,d\mu(x) \quad(A\in\mathcal{A}). \end{align*} $$
Let
$Z_n^Y(x)$
denote the last time the orbit
$\{T^k x\}_{k\geq 0}$
arrives in Y until time n, that is,
$$ \begin{align*} Z_n^Y(x) = \max\{k\leq n: T^k x \in Y\} \quad (n\in\mathbb{N}_0,\;x\in X). \end{align*} $$
Theorem 3.1. Suppose the following conditions are satisfied:
-
(A1)
$w_n^Y\sim n^{1-\alpha }\ell (n) (n\to \infty )$
for some
$\alpha \in (0,1)$
and some positive, measurable function
$\ell :(0,\infty )\to (0,\infty )$
slowly varying at
$\infty $
; -
(A2) there exists
$N\in \mathbb {N}$
such that
$ \{({1}/{w_n^{Y}})\sum _{k=0}^{n-1}\widehat {T}^k 1_{Y_k}\}_{n\geq N}$
is uniformly sweeping for
$1_Y$
; -
(A3) there exists
$H\in L^\infty (\mu )$
such that
$$ \begin{align*} \lim_{n\to\infty}\frac{1}{w_n^Y} \sum_{k=0}^{n-1} \widehat{T}^k 1_{Y_k} = H \quad \text{in }L^\infty(\mu). \end{align*} $$
Let
$\{c(n)\}_{n\geq 0}$
be a positive sequence satisfying
$$ \begin{align} c(n)\to0, \quad c(n)n\to\infty \quad(n\to\infty). \end{align} $$
Then,
$$ \begin{align} \mu_{H}\bigg(\frac{Z^Y_n}{n} \leq c(n) \bigg) \sim \frac{\sin(\pi\alpha)}{\pi\alpha} \frac{c(n)^\alpha \ell(n)}{\ell(c(n)n)} \quad(n\to\infty). \end{align} $$
The proof of Theorem 3.1 will be given in §5.
Remark 3.2. Under the setting of Theorem 3.1, fix
$\varepsilon \in (0,\alpha )$
arbitrarily. Then, the Potter bounds for slowly varying functions [Reference Bingham, Goldie and Teugels6, Theorem 1.5.6] implies that there exist
$C_\varepsilon \geq 1$
and
$N_\varepsilon \in \mathbb {N}$
such that, for any
$n\geq N_\varepsilon $
, we have
$c(n)\leq 1$
and
$$ \begin{align*} C_\varepsilon^{-1} c(n)^\varepsilon \leq \frac{\ell(n)}{\ell(c(n)n)}\leq C_\varepsilon c(n)^{-\varepsilon}. \end{align*} $$
Thus, the right-hand side of (3.2) converges to
$0$
as
$n\to \infty $
.
Remark 3.3. If we further assume
$$ \begin{align} \frac{\ell(n)}{\ell(c(n)n)}\to 1 \quad(n\to\infty), \end{align} $$
then we obtain
$$ \begin{align} \mu_{H}\bigg(\frac{Z^Y_n}{n} \leq c(n) \bigg) \sim \frac{\sin(\pi\alpha)}{\pi\alpha} c(n)^\alpha \quad(n\to\infty). \end{align} $$
Remark 3.4. Fix any positive, measurable function
$\ell :(0,\infty )\to (0,\infty )$
slowly varying at
$\infty $
. Then, there exists a non-increasing, positive sequence
$\{c(n)\}_{n\geq 0}$
satisfying (3.1) and (3.3). Indeed, we use the uniform convergence theorem for slowly varying functions [Reference Bingham, Goldie and Teugels6, Theorem 1.2.1] to take a strictly increasing sequence
$\{M_N\}_{N\geq 1}\subset \mathbb {N}$
so that
$$ \begin{align*} \sup \bigg\{\bigg|\frac{\ell(t)}{\ell(\unicode{x3bb} t)}-1\bigg|:\unicode{x3bb}\in[N^{-1}, 1], \;t\geq M_N \bigg\} \leq \frac{1}{N} \quad(N\in\mathbb{N}). \end{align*} $$
Set
$c(n)=1$
for
$0\leq n < M_1$
and
$c(n)=N^{-1/2}$
for
$M_N\leq n < M_{N+1} (N\in \mathbb {N})$
. It is easy to check that
$\{c(n)\}_{n\geq 0 }$
satisfies (3.1) and (3.3).
Remark 3.5. (Comparison with the Dynkin–Lamperti generalized arcsine law)
Let us recall the Dynkin–Lamperti generalized arcsine law for waiting times. Assume conditions (A1) and (A3) of Theorem 3.1 are fulfilled. Then, for any
$\mu $
-absolutely continuous probability measure
$\nu $
on
$(X,\mathcal {A})$
and any
$0\leq t\leq 1$
, we have
$$ \begin{align} \lim_{n\to\infty}\nu\bigg(\frac{Z^Y_n}{n} \leq t \bigg) = \frac{\sin(\pi\alpha)}{\pi}\int_0^t \frac{ds}{s^{1-\alpha}(1-s)^{\alpha}}, \end{align} $$
which follows from [Reference Zweimüller33, Theorem 2.3]. See also [Reference Thaler and Zweimüller27, Theorem 3.3] and [Reference Kocheim and Zweimüller11, Theorem 2.1]. The limit is the distribution function of the
$\mathrm {Beta}(\alpha ,1-\alpha )$
-distribution. In the case where
$\alpha =1/2$
, this distribution is the usual arcsine distribution. We emphasize that the right-hand side of (3.5) does not depend on the choice of
$\nu $
because of the ergodicity of T. Note that the right-hand side of (3.4) is asymptotically the same as the right-hand side of (3.5) with
$t=c(n)$
, as
$n\to \infty $
. Nevertheless, (3.4) does not follow from (3.5) directly. We do not know whether (3.2) remains valid in the case
$\mu _H$
is replaced by other suitable probability measures
$\nu $
except for
$\mu _{\widehat {T}^k H}$
(see also Corollaries 3.6 and 3.7, Theorem 3.8, and Remark 3.9). The difficulty is that the
$L^1$
-characterization of the ergodicity [Reference Zweimüller33, Theorem 3.1] is inadequate for this purpose, although it is significant for (3.5).
In the following two corollaries, we will consider what happens when we replace
$\mu _{H}$
in the left-hand side of (3.2) by other finite measures.
Corollary 3.6. Let
$k\in \mathbb {N}_{0}$
. Under the setting of Theorem 3.1,
$$ \begin{align*} \mu_{H}\bigg(\frac{Z^Y_n\circ T^k}{n} \leq c(n) \bigg) \bigg(= \mu_{\widehat{T}^k H}\bigg(\frac{Z^Y_n}{n} \leq c(n) \bigg) \bigg) \sim \frac{\sin(\pi\alpha)}{\pi\alpha}\frac{c(n)^\alpha \ell(n)}{\ell(c(n)n)} \quad(n\to\infty). \end{align*} $$
Proof of Corollary 3.6 by using Theorem 3.1
Note that
$Z^Y_{n}\circ T^k= \max \{0,\;Z^Y_{n+k}-k\}$
and hence,
$\{Z^Y_n\circ T^k \leq nc(n)\}=\{Z^Y_{n+k}\leq nc(n)+k\}$
. In addition,
$$ \begin{align*} \frac{nc(n)+k}{n+k} \to0 \quad \text{and} \quad nc(n)+k\to \infty (n\to\infty). \end{align*} $$
In other words, (3.1) is satisfied with n and
$c(n)$
replaced by
$n+k$
and
$(nc(n)+k)/(n+k)$
, respectively. Therefore, Theorem 3.1 yields
$$ \begin{align*} \mu_{H}\bigg(\frac{Z^Y_n\circ T^k}{n} \leq c(n) \bigg) &= \mu_{H}\bigg(\frac{Z^Y_{n+k}}{n}\leq c(n)+\frac{k}{n}\bigg) \\ &= \mu_H\bigg(\frac{Z^Y_{n+k}}{n+k}\leq \frac{nc(n)+k}{n+k}\bigg) \\ &\sim \frac{\sin(\pi\alpha)}{\pi\alpha} \bigg(\frac{c(n)n+k}{n+k}\bigg)^\alpha \frac{\ell(n+k)}{\ell(c(n)n+k)} \\ &\sim \frac{\sin(\pi\alpha)}{\pi\alpha}\frac{c(n)^\alpha \ell(n)}{\ell(c(n)n)} \quad (n\to\infty). \end{align*} $$
Here, we used the uniform convergence theorem for slowly varying functions. This completes the proof.
Corollary 3.7. Suppose that conditions (A1), (A2), (A3) in Theorem 3.1 are fulfilled. Let
$G\in \{u\in L^1(\mu ):u\geq 0\}$
. Then, the following assertions hold.
-
(1) Assume that G is uniformly sweeping for
$1_Y$
. Then, there exists
$C_1\in (0,\infty )$
such that, for any positive sequence
$\{c(n)\}_{n\geq 0}$
satisfying (3.1), we have (3.6)
$$ \begin{align} C_1 \leq \liminf_{n\to\infty} \frac{\ell(c(n)n)}{c(n)^\alpha \ell(n)} \mu_G\bigg(\frac{Z^Y_n}{n} \leq c(n) \bigg). \end{align} $$
-
(2) Assume that
$1_Y$
is uniformly sweeping for G. Then, there exists
$C_2\in (0,\infty )$
such that, for any positive sequence
$\{c(n)\}_{n\geq 0}$
satisfying (3.1), we have (3.7)
$$ \begin{align} \limsup_{n\to\infty} \frac{\ell(c(n)n)}{c(n)^\alpha \ell(n)} \mu_G\bigg(\frac{Z^Y_n}{n} \leq c(n) \bigg) \leq C_2. \end{align} $$
Proof of Corollary 3.7 by using Theorem 3.1
(1) By the assumption, G is also uniformly sweeping for H. Take
$K\in \mathbb {N}$
so large that
$\sum _{k=0}^{K-1} \widehat {T}^k G\geq K^{-1}H$
, a.e. Let
$k\in \{0,1,\ldots ,K\}$
. Note that
$Z^Y_{n}\circ T^k +K\geq Z^Y_{n}$
and hence,
$$ \begin{align*} \mu_G \bigg(\frac{Z^Y_{n}}{n}\leq c(n)\bigg) \geq \mu_G \bigg(\frac{Z^Y_{n}\circ T^k}{n}\leq c(n)-\frac{K}{n}\bigg). \end{align*} $$
Therefore, Theorem 3.1 yields that
$$ \begin{align*} \begin{split} \mu_G \bigg(\frac{Z^Y_{n}}{n}\leq c(n)\bigg) &\geq \frac{1}{K}\sum_{k=0}^{K-1} \mu_G \bigg(\frac{Z^Y_{n}\circ T^k}{n}\leq c(n)-\frac{K}{n}\bigg) \\ &\geq \frac{1}{K^2} \mu_{H} \bigg(\frac{Z^Y_{n}}{n}\leq c(n)-\frac{K}{n}\bigg) \\ &\sim \frac{\sin(\pi\alpha)}{K^2\pi\alpha}\frac{c(n)^\alpha \ell(n)}{\ell(c(n)n)} \quad (n\to\infty), \end{split} \end{align*} $$
which implies the desired result.
(2) By the assumption, H is also uniformly sweeping for G. Take
$K\in \mathbb {N}$
so large that
$G\leq K\sum _{k=0}^{K-1} \widehat {T}^k H$
, a.e. Then, we use Corollary 3.6 to obtain
$$ \begin{align*} \begin{split} \mu_G \bigg(\frac{Z^Y_{n}}{n}\leq c(n)\bigg) &\leq K \sum_{k=0}^{K-1} \mu_{H} \bigg(\frac{Z^Y_n\circ T^k}{n}\leq c(n)\bigg) \\ &\sim K^2 \frac{\sin(\pi\alpha)}{\pi\alpha}\frac{c(n)^\alpha \ell(n)}{\ell(c(n)n)} \quad(n\to\infty), \end{split} \end{align*} $$
as desired.
We will also give the proof of the following theorem in §5.
Theorem 3.8. Suppose that conditions (A1) and (A2) of Theorem 3.1 are fulfilled. Let
$G\in \{u\in L^1(\mu ):u\geq 0\}$
. Then, assertion (2) of Corollary 3.7 holds.
In other words, assertion (2) of Corollary 3.7 remains valid without assuming the existence of the asymptotic entrance density H. The reader may expect assertion (1) also remains valid under a similar setting, but we do not know whether it is true. The reason is that
$\mu _G(Z_n^Y/n\leq c(n))$
can be bounded above but not below by double Laplace transform of
$Z_n^Y$
, as we shall see in the proof of Theorem 3.8.
Remark 3.9. Let
$\alpha \in (0,1)$
and let
$\{c(n)\}_{n\geq 0}$
be a non-increasing,
$(0,1]$
-valued sequence satisfying
$c(0)=1$
and
$c(n)\to 0 (n\to \infty )$
. Then, there exists a
$\mu $
-probability density function G such that
$$ \begin{align} \limsup_{n\to\infty}\frac{1}{c(n)^\alpha}\mu_{G}\bigg(\frac{Z_n^Y}{n}\leq c(n)\bigg)=\infty, \end{align} $$
as we shall see below. Indeed, let
$$ \begin{align*} N_0=0\quad \text{and}\quad N_{k}=\min\{n> N_{k-1}: \mu(Y\cap\{\varphi=n\})>0\} \quad(k\in \mathbb{N}). \end{align*} $$
Then,
$\{N_k\}_{k\geq 0}\subset \mathbb {N}_0$
is strictly increasing. We define the
$\mu $
-probability density function
$G:X\to [0,\infty )$
as
$$ \begin{align*} G= \begin{cases} (c(N_{k-1})^{\alpha/2}-c(N_{k})^{\alpha/2})/\mu(Y\cap\{\varphi=N_k\}) &\text{on }Y\cap \{\varphi=N_k\} (k\in\mathbb{N}), \\ 0 &\text{otherwise.} \end{cases} \end{align*} $$
Then,
$ \mu _G(\varphi> N_k)=c(N_k)^{\alpha /2}$
and hence,
$$ \begin{align*} \frac{1}{c(N_k)^\alpha} \mu_G\bigg(\frac{Z^Y_{N_k}}{N_k} \leq c(N_k) \bigg) \geq \frac{1}{c(N_k)^\alpha} \mu_G(\varphi> N_k) = c(N_k)^{-\alpha/2} \to\infty\quad(k\to\infty), \end{align*} $$
which implies (3.8).
3.2 Large deviation estimates related to the Darling–Kac theorem
For
$A\in \mathcal {A}$
, let
$S_n^{A}(x)$
denote the occupation time in A of the orbit
$\{T^k x\}_{k\geq 0}$
from time
$1$
to time n, that is,
$$ \begin{align*} S_n^{A}(x)=\sum_{k=1}^n 1_{A}(T^k x)\quad (n\in\mathbb{N}_0,\;x\in X). \end{align*} $$
In the following, we consider occupation times in a set of finite measure.
Theorem 3.10. Suppose the following conditions are satisfied:
-
(B1)
$w_n^Y\sim n^{1-\alpha }\ell (n) (n\to \infty )$
for some
$\alpha \in (0,1)$
and some positive, measurable function
$\ell :(0,\infty )\to (0,\infty )$
slowly varying at
$\infty $
; -
(B2)
$\{({1}/{w^{Y}_n})\sum _{k=0}^{n-1}\widehat {T}^k 1_{Y_k}\}_{n\geq 1}$
is
$L^\infty (\mu )$
-bounded.
For
$t>0$
, set
$$ \begin{align*} a(t) = \frac{t}{\Gamma(1+\alpha)Q^Y(t^{-1})} \sim \frac{t^{\alpha}}{\Gamma(1+\alpha)\Gamma(2-\alpha)\ell(t)} \quad (t\to\infty). \end{align*} $$
Let
$\{\widetilde {c}(n)\}_{n\geq 0}$
be a positive sequence satisfying
$$ \begin{align} \widetilde{c}(n)\to0\quad\text{and}\quad \widetilde{c}(n)a(n)\to\infty \quad(n\to\infty). \end{align} $$
Then,
$$ \begin{align} \mu_{1_Y}\bigg(\frac{S^{Y}_n}{a(n)} \leq \widetilde{c}(n) \bigg) \sim \frac{\sin(\pi\alpha)}{\pi\alpha}\widetilde{c}(n) \quad(n\to\infty). \end{align} $$
The proof of Theorem 3.10 will be given in §6.
Remark 3.11. (Comparison with the Darling–Kac theorem)
Let us recall the Darling–Kac theorem. Set
$$ \begin{align*} F(t)=\frac{1}{\pi\alpha}\int_0^t \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k!} \sin(\pi\alpha k)\Gamma(1+\alpha k)s^{k-1}\,ds \quad (t\geq0), \end{align*} $$
which is the distribution function of the Mittag–Leffler distribution of order
$\alpha $
with Laplace transform
$$ \begin{align*} \int_0^\infty e^{-\unicode{x3bb} t}\,dF(t) = \sum_{k=0}^\infty\frac{(-\unicode{x3bb})^k}{\Gamma(1+\alpha k)}\quad (\unicode{x3bb}\in \mathbb{R}). \end{align*} $$
See [Reference Darling and Kac7, Reference Pollard15] for the details. As a special case, the Mittag–Leffler distribution of order
$1/2$
is the half-normal distribution with mean
$2/\sqrt {\pi }$
. Suppose conditions (A1), (A2), (A3) of Theorem 3.1 are satisfied. Then, for any
$\mu $
-absolutely continuous probability measure
$\nu $
on X and for any
$t\geq 0$
, we have
$$ \begin{align} \lim_{n\to\infty}\nu\bigg(\frac{S^{Y}_n}{a(n)} \leq t \bigg) = F\bigg(\frac{t}{\Gamma(1+\alpha)\mu(Y)}\bigg), \end{align} $$
which follows from [Reference Thaler and Zweimüller27, Theorem 3.1]. See also [Reference Zweimüller33, Theorem 2.1] and [Reference Kocheim and Zweimüller11, Theorem 2.1]. Note that
$$ \begin{align*} F\bigg(\frac{\widetilde{c}(n)}{\Gamma(1+\alpha)\mu(Y)}\bigg) \sim \frac{\sin(\pi\alpha)}{\pi\alpha}\frac{\widetilde{c}(n)}{\mu(Y)} \quad (n\to\infty). \end{align*} $$
Corollary 3.12. Let
$k\in \mathbb {N}_{0}$
. Under the setting of Theorem 3.10,
$$ \begin{align*} \mu_{1_Y}\bigg(\frac{S^{Y}_n\circ T^k}{a(n)} \leq \widetilde{c}(n) \bigg) \bigg(= \mu_{\widehat{T}^k 1_Y}\bigg(\frac{S^{Y}_n}{a(n)} \leq \widetilde{c}(n) \bigg) \bigg) \sim \frac{\sin(\pi\alpha)}{\pi\alpha}\widetilde{c}(n) \quad (n\to\infty). \end{align*} $$
Proof of Corollary 3.12 by using Theorem 3.10
Since
$|S^Y_n-S^Y_n\circ T^k|\leq k$
, we see that
$$ \begin{align*} \mu_{1_Y}\bigg(\frac{S^{Y}_n}{a(n)} \leq \widetilde{c}(n)-\frac{k}{a(n)}\bigg) \leq \mu_{1_Y}\bigg(\frac{S^{Y}_n\circ T^k}{a(n)} \leq \widetilde{c}(n) \bigg) \leq \mu_{1_Y}\bigg(\frac{S^{Y}_n}{a(n)} \leq \widetilde{c}(n) +\frac{k}{a(n)} \bigg). \end{align*} $$
By Theorem 3.10,
$$ \begin{align*} \mu_{1_Y}\bigg(\frac{S^{Y}_n}{a(n)} \leq \widetilde{c}(n) \pm\frac{k}{a(n)} \bigg)\sim \frac{\sin(\pi\alpha)}{\pi\alpha}\widetilde{c}(n) \quad(n\to\infty), \end{align*} $$
as desired.
Corollary 3.13. Suppose that conditions (B1) and (B2) of Theorem 3.10 are fulfilled. Let
$G\in \{u\in L^1(\mu ):u\geq 0\}$
. Then, the following assertions hold.
-
(1) Assume that G is uniformly sweeping for
$1_Y$
. Then, there exists
$C_1\in (0,\infty )$
such that, for any positive sequence
$\{\widetilde {c}(n)\}_{n\geq 0}$
satisfying (3.9), we have (3.12)
$$ \begin{align} C_1 \leq \liminf_{n\to\infty}\frac{1}{\widetilde{c}(n)}\mu_G\bigg(\frac{S_n^Y}{a(n)} \leq \widetilde{c} (n) \bigg). \end{align} $$
-
(2) Assume that
$1_Y$
is uniformly sweeping for G. Then, there exists
$C_2\in (0,\infty )$
such that, for any positive sequence
$\{\widetilde {c}(n)\}_{n\geq 0}$
satisfying (3.9), we have (3.13)
$$ \begin{align} \limsup_{n\to\infty}\frac{1}{\widetilde{c}(n)}\mu_G\bigg(\frac{S_n^{Y}}{a(n)} \leq \widetilde{c}(n)\bigg) \leq C_2. \end{align} $$
Proof of Corollary 3.13 by using Theorem 3.10
(1) Take
$K\in \mathbb {N}$
so large that
$\sum _{k=0}^{K-1} \widehat {T}^k G\geq K^{-1}1_Y$
, a.e. Let
$k\in \{0,1,2,\ldots ,K\}$
. Then,
$S_n^Y\leq S_n^Y\circ T^k +k\leq S_n^Y\circ T^k +K$
and hence,
$$ \begin{align*} \mu_G\bigg(\frac{S^{Y}_n}{a(n)} \leq \widetilde{c}(n) \bigg) \geq \mu_G\bigg(\frac{S^{Y}_n\circ T^k}{a(n)} \leq \widetilde{c}(n)-\frac{K}{a(n)}\bigg). \end{align*} $$
Note that
$$ \begin{align*} \widetilde{c}(n)-\frac{K}{a(n)}\to0 \quad \text{and} \quad \bigg(\widetilde{c}(n)-\frac{K}{a(n)}\bigg)a(n)\to\infty \quad(n\to\infty). \end{align*} $$
Hence, Theorem 3.10 with
$\widetilde {c}(n)$
replaced by
$\widetilde {c}(n)-(K/a(n))$
implies
$$ \begin{align*} \begin{split} \mu_G\bigg(\frac{S^{Y}_n}{a(n)} \leq \widetilde{c}(n) \bigg) &\geq K^{-1}\sum_{k=0}^{K-1} \mu_G\bigg(\frac{S^{Y}_n\circ T^k}{a(n)} \leq \widetilde{c}(n)-\frac{K}{a(n)}\bigg) \\ &\geq K^{-2} \mu_{1_Y}\bigg(\frac{S^{Y}_n}{a(n)} \leq \widetilde{c}(n)-\frac{K}{a(n)}\bigg) \\ &\sim \frac{\sin(\pi\alpha)}{K^{2}\pi\alpha}\widetilde{c}(n) \quad(n\to\infty), \end{split} \end{align*} $$
which implies the desired result.
(2) Take
$K\in \mathbb {N}$
large enough that
$G\leq K\sum _{k=0}^{K-1} \widehat {T}^k 1_Y$
, a.e. Then, we use Corollary 3.12 to obtain
$$ \begin{align*} \begin{split} \mu_G\bigg(\frac{S^{Y}_n}{a(n)} \leq \widetilde{c}(n) \bigg) &\leq K\sum_{k=0}^{K-1}\mu_{1_Y}\bigg(\frac{S^{Y}_n\circ T^k}{a(n)} \leq \widetilde{c}(n)\bigg) \\ &\sim K^2\frac{\sin(\pi\alpha)}{\pi\alpha}\widetilde{c}(n) \quad(n\to\infty), \end{split} \end{align*} $$
as desired.
3.3 Large deviation estimates related to the Lamperti generalized arcsine law
In the following, we consider occupation times in sets of infinite measure under certain additional assumptions. Fix disjoint sets
$Y, A_1, A_2,\ldots ,A_d \in \mathcal {A}$
with
$d\in \mathbb {N}$
,
$d\geq 2$
,
$X=Y\cup \bigcup _{i=1}^d A_i$
,
$0<\mu (Y)<\infty $
, and
$\mu (A_i)=\infty (i=1,2,\ldots ,d)$
. We assume Y dynamically separates
$A_1, A_2,\ldots ,A_d$
(under the action of T), that is,
$A_i\cap T^{-1}A_j =\emptyset $
whenever
$i\neq j$
. Then, the condition [
$x\in A_i$
and
$T^n x\in A_j (i\neq j)$
] implies
$n\geq 2$
and the existence of
$k=k(x)\in \{1,\ldots ,n-1\}$
for which
$T^k x\in Y$
. As shown in [Reference Thaler and Zweimüller27, (6.6)],
$$ \begin{align} 1_{Y_n\cap A_i} = \sum_{k>n}\widehat{T}^{k-n}1_{Y\cap (T^{-1}A_i) \cap\{\varphi=k\}} \quad \text{a.e.}\quad(n\in\mathbb{N}, \; i=1,2,\ldots,d), \end{align} $$
and
$\mu (Y_n\cap A_i)=\mu (Y\cap (T^{-1}A_i) \cap \{\varphi>n\})$
(
$n\in \mathbb {N}$
,
$i=1,2,\ldots ,d$
). Let
$\{w^{Y, A_i}_n\}_{n\geq 0}$
denote the wandering rate of Y starting from
$A_i$
, which is given by
$$ \begin{align} w^{Y, A_i}_n &= \mu\bigg( \bigcup_{k=0}^{n-1} (T^{-k}Y)\cap A_i \bigg) = \sum_{k=0}^{n-1} \mu(Y_k\cap A_i) = \sum_{k=0}^{n-1} \int_Y \widehat{T}^k 1_{Y_k\cap A_i} \,d\mu \nonumber\\ &= \sum_{k=1}^{n-1} \mu(Y\cap (T^{-1}A_i) \cap \{\varphi>k\}) \quad (n\in\mathbb{N}_0,\;i=1,2,\ldots,d). \end{align} $$
Since T is a CEMPT, we see
$\bigcup _{n\geq 0}(T^{-n}Y)\cap A_i=A_i$
, a.e. and hence,
$w^{Y,A_i}_n\to \infty ({n\to \infty },\,i=1,\ldots ,d)$
. We write
$Q^{Y, A_i}(s)$
for Laplace transform of
$\{w^{Y,A_i}_{n+1}-w^{Y, A_i}_n\}_{n\geq 0}$
:
$$ \begin{align*} Q^{Y,A_i}(s) &=\sum_{n\geq0} e^{-ns}(w_{n+1}^{Y,A_i}-w_{n}^{Y,A_i}) \\ &=\sum_{n\geq1}e^{-ns}\mu(Y\cap (T^{-1}A_i)\cap\{\varphi>n\})\quad(s>0,\;i=1,2,\ldots,d). \end{align*} $$
Then,
$w^{Y}_n=\mu (Y)+\sum _{i=1}^d w^{Y, A_i}_n$
and
$Q^Y(s)=\mu (Y)+\sum _{i=1}^d Q^{Y,A_i}(s)$
. In addition,
$0<Q^{Y, A_i}(s)<\infty $
and
$Q^{Y,A_i}(s)\to \infty (s\to 0+,\, i=1,\ldots ,d)$
. Let
$\alpha , \beta _1,\beta _2,\ldots ,\beta _d\in (0,1)$
with
$\sum _{i=1}^d \beta _i=1$
and let
$\ell :(0,\infty )\to (0,\infty )$
be a positive, measurable function slowly varying at
$\infty $
. By Karamata’s Tauberian theorem, the condition
$$ \begin{align*} w_n^{Y,A_i} \sim \beta_i n^{1-\alpha}\ell(n) \quad (n\to\infty,\;i=1,2,\ldots,d) \end{align*} $$
is equivalent to
$$ \begin{align} Q^{Y,A_i}(s)\sim \Gamma(2-\alpha) \beta_i s^{-1+\alpha} \ell(s^{-1}) \quad (s\to 0+,\;i=1,2,\ldots,d). \end{align} $$
The following lemma will be used in §8.
Lemma 3.14. Fix
$i\in \{1,2,\ldots ,d\}$
. Assume that there exists
$H^{(i)}\in L^\infty (\mu )$
such that
$$ \begin{align*} &\|\widehat{T}^n 1_{Y\cap(T^{-1}A_i)\cap \{\varphi=n\}} - \mu(Y\cap(T^{-1}A_i)\cap \{\varphi=n\})H^{(i)} \|_{L^\infty(\mu)} \\ &\quad= o(\mu(Y\cap(T^{-1}A_i)\cap \{\varphi=n\})) \quad (n\to\infty). \end{align*} $$
Then,
$$ \begin{align*} \lim_{n\to\infty}\bigg(\frac{1}{w_n^{Y,A_i}} \sum_{k=0}^{n-1} \widehat{T}^k 1_{Y_k\cap A_i}\bigg) = H^{(i)} \quad \text{in }L^\infty(\mu). \end{align*} $$
We omit the proof of Lemma 3.14, since it is almost the same as that of Lemma 2.1.
Theorem 3.15. Suppose the following conditions are satisfied.
-
(C1) For
$d\in \mathbb {N}$
with
$d\geq 2$
, let
$A_1,\ldots ,A_d\in \mathcal {A}$
with
$X=Y\cup \bigcup _{i=1}^d A_i$
and
$\mu (A_i)=\infty (i=1,2,\ldots ,d)$
. In addition,
$Y, A_1, A_2,\ldots ,A_d$
are disjoint sets, and Y dynamically separates
$A_1, A_2,\ldots ,A_d$
. -
(C2)
$w^{Y, A_i}_n\sim \beta _i n^{1-\alpha }\ell (n) (n\to \infty ,\; i=1,2,\ldots ,d)$
for some
$\alpha , \beta _1, \beta _2,\ldots ,\beta _d \in (0,1)$
with
$\sum _{i=1}^d \beta _i=1$
and for some positive, measurable function
$\ell :(0,\infty )\to (0,\infty )$
slowly varying at
$\infty $
. -
(C3)
$ \{ ({1}/{w_n^{Y,A_d}}) \sum _{k=0}^{n-1}\widehat {T}^k 1_{Y_k\cap A_d} \}_{n\geq 2}$
is
$L^\infty (\mu )$
-bounded. -
(C4) There exists
$N\geq 2$
such that
$ \{{1}/{w_n^{Y,A_i}}\sum _{k=0}^{n-1}\widehat {T}^k 1_{Y_k\cap A_i}\}_{n\geq N,\;i=1,\ldots ,d-1}$
is uniformly sweeping for
$1_Y$
. -
(C5) There exist
$H^{(1)},\ldots ,H^{(d-1)}\in L^\infty (\mu )$
such that
$$ \begin{align*} \lim_{n\to\infty}\bigg(\frac{1}{w_n^{Y,A_i}} \sum_{k=0}^{n-1} \widehat{T}^k 1_{Y_k\cap A_i}\bigg) = H^{(i)} \quad \text{in }L^\infty(\mu) \quad(i=1,\ldots,d-1). \end{align*} $$
Let
$\unicode{x3bb} _1,\ldots ,\unicode{x3bb} _{d-1}\in (0,\infty )$
and let
$\{c(n)\}_{n\geq 0}$
be a positive sequence satisfying (3.1). Set
$\widetilde {H}=\sum _{i=1}^{d-1}\beta _i \unicode{x3bb} _i^\alpha H^{(i)}$
. Then,
The proof of Theorem 3.15 will be given in §7.
Remark 3.16. Under the setting of Theorem 3.15 with
$d=2$
and
$\unicode{x3bb} _{1}=1$
, let us assume (3.3). Then, we obtain
$$ \begin{align} \mu_{H^{(1)}}\bigg(\frac{S_n^{A_1}}{n}\leq c(n)\bigg) \sim \frac{\beta_2}{\beta_1} \frac{\sin(\pi \alpha)}{\pi\alpha} c(n)^\alpha \quad (n\to\infty). \end{align} $$
Remark 3.17. If
$\unicode{x3bb} _i=0$
for some i, then (3.17) does not remain valid. For example, let
$d\geq 3$
,
$\unicode{x3bb} _1, \ldots , \unicode{x3bb} _{d-2}\in (0,\infty )$
, and
$\unicode{x3bb} _{d-1}=0$
. Then,
which follows from Theorem 3.15.
Remark 3.18. (Comparison with the Lamperti generalized arcsine law)
Let us recall the Lamperti generalized arcsine law for occupation times. Suppose conditions (C1), (C2), (C5) of Theorem 3.15 and condition (A2) of Theorem 3.1 are fulfilled with
$d=2$
. Set
$b=\beta _2/\beta _1$
. Then, for any
$\mu $
-absolutely continuous probability measure
$\nu $
on
$(X,\mathcal {A})$
and for any
$0\leq t\leq 1$
, we have
$$ \begin{align} \lim_{n\to\infty}\nu\bigg(\frac{S_n^{A_1}}{n} \leq t\bigg) &= \frac{b\sin(\pi\alpha)}{\pi} \int_0^t \frac{s^{\alpha-1} (1-s)^{\alpha-1}\,ds} {b^2s^{2\alpha} +2bs^\alpha (1-s)^\alpha \cos(\pi\alpha) +(1-s)^{2\alpha}} \nonumber\\ &= \frac{1}{\pi\alpha} \mathrm{arccot}\bigg(\frac{(1-t)^\alpha}{b \sin(\pi\alpha) t^\alpha}+\cot(\pi\alpha)\bigg), \end{align} $$
as shown in [Reference Zweimüller33, Theorem 2.2]. See also [Reference Thaler and Zweimüller27, Theorem 3.2] and [Reference Sera and Yano18, Theorem 2.7]. The limit is the distribution function of the Lamperti generalized arcsine distribution of parameter
$(\alpha ,\beta _1)$
. In the case where
$\alpha =\beta _1=\beta _2=1/2$
, this distribution is the usual arcsine distribution. Note that the right-hand side of (3.18) is asymptotically the same as the right-hand side of (3.19) with
$t=c(n)$
. Nevertheless, (3.18) does not follow from (3.19) directly.
The proofs of the following two corollaries are almost the same as those of Corollaries 3.6, 3.7, 3.12, and 3.13, so we omit them.
Corollary 3.19. Let
$k\in \mathbb {N}_0$
. Under the setting of Theorem 3.15,
Corollary 3.20. Suppose that conditions (C1)–(C5) of Theorem 3.15 are satisfied. Let
$G\in \{u\in L^1(\mu ):u\geq 0\}$
and
$\unicode{x3bb} _1,\ldots ,\unicode{x3bb} _{d-1}\in (0,\infty )$
. Then, the following assertions hold.
-
(1) Assume that G is uniformly sweeping for
$1_Y$
. Then, there exists
$C_1\in (0,\infty )$
such that, for any positive sequence
$\{c(n)\}_{n\geq 0}$
satisfying (3.1), we have -
(2) Assume that
$1_Y$
is uniformly sweeping for G. Then, there exists
$C_2\in (0,\infty )$
such that, for any positive sequence
$\{c(n)\}_{n\geq 0}$
satisfying (3.1), we have
Assertion (2) remains valid even if we drop condition (C5) of Theorem 3.15.
4 Analytical tools
In this section, we prove lemmas needed in the following.
Lemma 4.1. Let
$f_n:(0,\infty )\to [0,\infty ) (n\in \mathbb {N}\cup \{\infty \})$
be non-increasing functions. Assume there exists a non-empty open interval
$I\subset (0,\infty )$
such that for any
$q\in I$
,
$$ \begin{align*} \lim_{n\to\infty}\int_0^\infty e^{-qu}f_n(u)\,du = \int_0^\infty e^{-qu}f_\infty(u)\,du<\infty. \end{align*} $$
Then,
$\lim _{n\to \infty }f_n(u)= f_\infty (u)$
for all continuity points
$u\in (0,\infty )$
of
$f_\infty $
.
See [Reference Sera and Yano18, Lemma 3.2] for the proof of Lemma 4.1.
Lemma 4.2. Fix a constant
$C>0$
. Let
$S_n:X\to [0,Cn] (n\in \mathbb {N}_0)$
be measurable functions, and let
$\unicode{x3bb} :(0,\infty )\to [0,\infty )$
be a non-negative function with
$\unicode{x3bb} (t)\to 0 (t\to \infty )$
. Suppose
$\nu _t (t>0)$
are non-zero finite measures on X. Then, for any
$q>0$
,
$$ \begin{align} &\sum_{n=0}^\infty e^{-nq t^{-1}} \int_X \exp(-\unicode{x3bb}(t) S_{n})\, d\nu_t \nonumber\\[5pt] &\quad\sim t\int_0^\infty e^{-qu}\bigg(\int_X \exp(-\unicode{x3bb}(t) S_{[ut]})\,d\nu_t\bigg)\,du \quad(t\to\infty). \end{align} $$
Proof. For fixed
$q>0$
, let
$l(t)$
and
$r(t)$
denote the left-hand and right-hand sides of (4.1), respectively. Note that
$$ \begin{align} r(t) = \sum_{n=0}^\infty t \int_{nt^{-1}}^{(n+1)t^{-1}} e^{-qu} \bigg(\int_X \exp(-\unicode{x3bb}(t)S_{n})\,d\nu_t\bigg) \,du, \end{align} $$
and hence,
$$ \begin{align*} \begin{split} 0 \leq l(t) - r(t) \leq \sum_{n=0}^\infty (e^{-nq t^{-1}}-e^{-(n+1)q t^{-1}})\nu_t(X) = \nu_t(X). \end{split} \end{align*} $$
In addition, since
$0\leq S_{[ut]}\leq Cut$
, we have
$$ \begin{align*} r(t) \geq \nu_t(X)t\int_0^\infty \exp(-(q+C\unicode{x3bb}(t)t) u)\,du = \frac{\nu_t(X)}{qt^{-1}+C\unicode{x3bb}(t)}. \end{align*} $$
Therefore, we obtain
$$ \begin{align*} 1 \leq \frac{l(t)} {r(t)} \leq 1+qt^{-1}+C\unicode{x3bb}(t) \to 1 \quad (t\to\infty). \end{align*} $$
This completes the proof.
The following three lemmas are slight extensions of [Reference Thaler and Zweimüller27, Lemma 4.2].
Lemma 4.3. Fix
$t_0>0$
and
$K\in \mathbb {N}_0$
. Suppose that the following conditions are fulfilled.
-
(i)
$\{H_t\}_{t\geq t_0}\cup \{G\}\subset \{u\in L^1(\mu ):u\geq 0\}$
. In addition,
$\{H_t\}_{t\geq t_0}$
is uniformly sweeping in K steps for G. -
(ii)
$R_{n,t}:X\to (0,1] (n\in \mathbb {N}_0,\;t>0)$
are measurable functions with
$$ \begin{align*} \sup\bigg\{\bigg\|\frac{R_{n,t}\circ T^k}{R_{n+k,t}}\bigg\|_{L^\infty(\mu)}:n,k\in \mathbb{N}_0,\;0\leq k\leq K,\; t\geq t_0\bigg\}<\infty. \end{align*} $$
Then, for any
$q>0$
, we have
$$ \begin{align*} \sup_{t\geq t_0} \frac { \sum_{n\geq0} e^{-nqt^{-1}} \int_X R_{n,t}G\,d\mu } { \sum_{n\geq0} e^{-nqt^{-1}} \int_X R_{n,t} H_t \,d\mu} <\infty. \end{align*} $$
Proof. By condition (i), there exists
$C_0>0$
such that
$G\leq C_0\sum _{k=0}^K \widehat {T}^k H_t$
a.e. for any
$t\geq t_0$
. Moreover, by condition (ii), we can take
$C\geq C_0$
large enough so that
$R_{n,t}\circ T^k\leq CR_{n+k,t}$
a.e. for any
$n,k\in \mathbb {N}_0$
with
$0\leq k\leq K$
and for any
$t\geq t_0$
. Then,
$$ \begin{align*} \int_X R_{n,t}G\,d\mu &\leq C\int_X R_{n,t}\bigg(\sum_{k=0}^K \widehat{T}^k H_t\bigg)\,d\mu = C\sum_{k=0}^K \int_X (R_{n,t}\circ T^k)H_t \,d\mu \\ &\leq C^2 \sum_{k=0}^K \int_X R_{n+k,t}H_t\,d\mu \quad(n\in\mathbb{N}_0,\;t\geq t_0). \end{align*} $$
This implies
$$ \begin{align*} \sum_{n\geq0} e^{-nqt^{-1}} \int_X R_{n,t}G\,d\mu & \leq C^2 \sum_{k=0}^K \sum_{n\geq k} e^{-(n-k)qt^{-1}} \int_X R_{n,t}H_t\,d\mu \\ &\leq \bigg( C^2\sum_{k=0}^{K}e^{kqt_0^{-1}} \bigg) \bigg( \sum_{n\geq0} e^{-nqt^{-1}} \int_X R_{n,t}H_t\,d\mu\bigg) \quad(t\geq t_0) , \end{align*} $$
which completes the proof.
Lemma 4.4. (Integrating transforms)
Under the assumptions of Lemma 4.3 with
$G=1_Y$
, we further suppose the following condition.
-
(iii)
$\{H_t\}_{t\geq t_0}\cup \{H\}\subset \{u\in L^\infty (\mu ):u\geq 0\; \text {and }u\text { is supported on }Y\}$
and
$H_t\to H$
in
$L^\infty (\mu ) (t\to \infty )$
.
Then, for any
$q>0$
,
$$ \begin{align*} \sum_{n\geq0}e^{-nqt^{-1}} \int_Y R_{n,t}H_t \,d\mu \sim \sum_{n\geq0} e^{-nqt^{-1}} \int_Y R_{n,t}H\,d\mu \quad(t\to\infty). \end{align*} $$
Proof. Note that
$H_t$
is supported on Y. By Lemma 4.3 with
$G=1_Y$
,
$$ \begin{align*} C:=\sup_{t\geq t_0}\frac { \sum_{n\geq0} e^{-nqt^{-1}} \int_Y R_{n,t}\,d\mu } { \sum_{n\geq0} e^{-nqt^{-1}} \int_Y R_{n,t}H_t \,d\mu} = \sup_{t\geq t_0}\frac { \sum_{n\geq0} e^{-nqt^{-1}} \int_X R_{n,t}1_Y\,d\mu } { \sum_{n\geq0} e^{-nqt^{-1}} \int_X R_{n,t}H_t \,d\mu} <\infty. \end{align*} $$
Therefore, for
$t\geq t_0$
,
$$ \begin{align*} \bigg| \frac {\sum_{n\geq0}e^{-nqt^{-1}} \int_Y R_{n,t}H \,d\mu} {\sum_{n\geq0}e^{-nqt^{-1}} \int_Y R_{n,t}H_t \,d\mu} -1\bigg| & = \frac { | \sum_{n\geq0} e^{-nqt^{-1}} \int_Y R_{n,t}(H-H_t)\,d\mu |} { \sum_{n\geq0} e^{-nqt^{-1}} \int_Y R_{n,t}H_t \,d\mu} \\[5pt] &\leq C \|H-H_t\|_{L^\infty(\mu)} \to0 \quad(t\to\infty), \end{align*} $$
as desired.
Lemma 4.5. Let
$\{v_n\}_{n\geq 0}\subset \{u\in L^\infty (\mu ):u\geq 0\;\text {and }u\text { is supported on }Y\}$
with
$$ \begin{align*} 0<\sum_{n\geq0}e^{-ns}\int_Y v_n\,d\mu <\infty \quad (s>0). \end{align*} $$
Let
$\unicode{x3bb} :(0,\infty )\to (0,\infty )$
be a positive function with
$\unicode{x3bb} (t)\to 0 (t\to \infty )$
. We define
$H_t\in L^1(\mu )$
as
$$ \begin{align} H_t = \frac {\sum_{n\geq0}e^{-n\unicode{x3bb}(t)} v_n } {\sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\int_Y v_n\,d\mu} \quad (t>0). \end{align} $$
Then, the following assertions hold.
-
(1)
$H_t$
can be represented as
$$ \begin{align*} H_t = \frac {\sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n v_k} {\sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n \int_Y v_k\,d\mu} \quad(t>0). \end{align*} $$
-
(2) Assume there exists
$N\in \mathbb {N}_0$
such that
$\sum _{k=0}^N \int _Y v_k\,d\mu>0$
and is uniformly sweeping in K steps for
$$ \begin{align*} \bigg\{\frac{\sum_{k=0}^n v_k}{\sum_{k=0}^n \int_Y v_k\,d\mu}\bigg\}_{n\geq N} \end{align*} $$
$1_Y$
. Then, there exists
$t_0>0$
such that
$\{H_t\}_{t\geq t_0}$
is uniformly sweeping in K steps for
$1_Y$
.
-
(3) Assume there exists
$C>0$
such that, for any
$n\in \mathbb {N}_0$
, (4.4)Then,
$$ \begin{align} \bigg\|\sum_{k=0}^n v_k\bigg\|_{L^\infty(\mu)} \leq C\sum_{k=0}^n \int_Y v_k\,d\mu. \end{align} $$
$\|H_t\|_{L^\infty (\mu )}\leq C$
for any
$t>0$
.
-
(4) Assume there exists
$H\in L^\infty (\mu )$
such that Then,
$$ \begin{align*} \frac{\sum_{k=0}^n v_k}{\sum_{k=0}^n \int_Y v_k\,d\mu} \to H \quad \text{in }L^\infty(\mu)\quad(n\to\infty). \end{align*} $$
$H_t\to H$
in
$L^\infty (\mu ) (t\to \infty )$
.
Proof. (1) By Fubini’s theorem,
$$ \begin{align*} \sum_{k\geq0} e^{-k\unicode{x3bb}(t)}v_k = \sum_{k\geq0} \bigg((1-e^{-\unicode{x3bb}(t)})\sum_{n\geq k}e^{-n\unicode{x3bb}(t)}\bigg)v_k = (1-e^{-\unicode{x3bb}(t)})\sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n v_k, \end{align*} $$
and hence,
$$ \begin{align*} H_t= \frac {\sum_{k\geq0}e^{-k\unicode{x3bb}(t)} v_k } {\sum_{k\geq0}e^{-k\unicode{x3bb}(t)}\int_Y v_k\,d\mu} = \frac {\sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n v_k} {\sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n \int_Y v_k\,d\mu}, \end{align*} $$
as desired.
(2) Take
$C>0$
large enough so that
$$ \begin{align*} C\sum_{m=0}^K T^m \bigg(\frac{\sum_{k=0}^n v_k}{\sum_{k=0}^n \int_Y v_k\,d\mu}\bigg) \geq 1_Y \quad\text{a.e., for any }n\geq N, \end{align*} $$
or, equivalently,
$$ \begin{align*} \sum_{k=0}^n \bigg(C\sum_{m=0}^K \widehat{T}^m v_k - \int_Y v_k\,d\mu \bigg) \geq 0 \quad \text{a.e.\, on }Y\text{, for any }n\geq N. \end{align*} $$
Then,
$$ \begin{align*} \sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n \bigg( C\sum_{m=0}^K\widehat{T}^m v_k - \int_Y v_k\,d\mu\bigg) & \geq \sum_{n=0}^{N-1}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n \bigg( C\sum_{m=0}^K\widehat{T}^m v_k - \int_Y v_k\,d\mu\bigg) \nonumber\\ &\geq -\sum_{n=0}^{N-1}\sum_{k=0}^n \int_Y v_k\,d\mu \quad \text{a.e.\, on }Y. \end{align*} $$
We have used the fact that
$\widehat {T}^m v_k\geq 0$
, a.e. Then,
$$ \begin{align} C\sum_{m=0}^K \widehat{T}^m H_t-1 &= \frac{\sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n ( C\sum_{m=0}^K\widehat{T}^m v_k - \int_Y v_k\,d\mu)} {\sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n \int_Y v_k\,d\mu} \nonumber\\[5pt] & \geq \frac{-\sum_{n=0}^{N-1}\sum_{k=0}^n \int_Y v_k\,d\mu} {\sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n \int_Y v_k\,d\mu} \quad \text{a.e.\, on }Y. \end{align} $$
Since
$\sum _{n\geq 0}e^{-n\unicode{x3bb} (t)}\sum _{k=0}^n \int _Y v_k\,d\mu \to \infty (t\to \infty )$
, we can take
$t_0>0$
large enough so that, for any
$t\geq t_0$
, the right-hand side of (4.5) is greater than
$-1/2$
. Thus, for any
$t\geq t_0$
, we have
$2C\sum _{m=0}^K \widehat {T}^m H_t\geq 1_Y$
a.e., and hence,
$\{H_t\}_{t\geq t_0}$
is uniformly sweeping in K steps for
$1_Y$
.
$$ \begin{align*} \|H_t\|_{L^\infty(\mu)} = \frac {\|\kern-2pt\sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n v_k\|_{L^\infty(\mu)}} {\sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n \int_Y v_k\,d\mu} \leq C, \end{align*} $$
as desired.
(4) Fix
$\varepsilon>0$
arbitrarily. Take
$N\in \mathbb {N}_0$
large enough so that
$\sum _{k=0}^N \int _Y v_k\,d\mu>0$
and
$$ \begin{align*} \bigg\|\frac{\sum_{k=0}^n v_k}{\sum_{k=0}^n \int_Y v_k\,d\mu} -H \bigg\|_{L^\infty(\mu)}\leq \varepsilon \quad \text{for any }n\geq N, \end{align*} $$
or, equivalently,
$$ \begin{align*} \bigg\|\sum_{k=0}^n \bigg(v_k-H \int_Y v_k\,d\mu\bigg)\bigg\|_{L^\infty(\mu)}\leq \varepsilon \sum_{k=0}^n \int_Y v_k\,d\mu \quad \text{for any }n\geq N. \end{align*} $$
Hence,
$$ \begin{align*} & \bigg\| \sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n \bigg( v_k - H\int_Y v_k\,d\mu\bigg) \bigg\|_{L^\infty(\mu)} \\ &\quad\leq \sum_{n=0}^{N-1} \sum_{k=0}^n \bigg\| v_k - H\int_Y v_k\,d\mu \bigg\|_{L^\infty(\mu)} + \varepsilon \sum_{n\geq N} e^{-n\unicode{x3bb}(t)} \sum_{k=0}^n\int_Y v_k\,d\mu, \end{align*} $$
which implies
$$ \begin{align*} \| H_t-H\|_{L^\infty(\mu)} & = \frac{ \|\kern-2pt \sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n ( v_k - H\int_Y v_k\,d\mu) \|_{L^\infty(\mu)} } {\sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n \int_Y v_k\,d\mu} \\[5pt] &\leq \frac{\sum_{n=0}^{N-1} \sum_{k=0}^n \| v_k - H\int_Y v_k\,d\mu \|_{L^\infty(\mu)}} {\sum_{n\geq0}e^{-n\unicode{x3bb}(t)}\sum_{k=0}^n \int_Y v_k\,d\mu}+\varepsilon \\[5pt] &\to\varepsilon \quad(t\to\infty). \end{align*} $$
This establishes the result, since
$\varepsilon>0$
was arbitrary.
The following lemma ensures that condition (ii) in Lemma 4.3 is satisfied for
$R_{n,t}=\exp (-\unicode{x3bb} (t) Z^Y_n)$
.
Lemma 4.6. Let
$\unicode{x3bb} :(0,\infty )\to [0,\infty )$
be a non-negative function with
$\unicode{x3bb} (t)\to 0 (t\to \infty )$
. Set
$$ \begin{align*} R_{n,t}=\exp(-\unicode{x3bb}(t)Z^Y_{n}) \quad(n\in\mathbb{N}_0,\;t>0). \end{align*} $$
Then, there exists a positive constant
$t_0>0$
such that for any
$n,k\in \mathbb {N}_0$
and
$t\geq t_0$
, we have
$$ \begin{align*} R_{n,t}\circ T^k \leq e^{k} R_{n+k,t}. \end{align*} $$
Proof. Take
$t_0>0$
so large that
$\unicode{x3bb} (t)\leq 1$
for any
$t\geq t_0$
. Since
$Z^Y_n\circ T^k\geq Z^Y_{n+k}-k$
, we have
$R_{n,t}\circ T^k \leq \exp (\unicode{x3bb} (t)k) R_{n+k,t}\leq e^k R_{n+k,t}$
for any
$t\geq t_0$
.
We can also prove the following lemma in almost the same way.
Lemma 4.7. Let
$d\in \mathbb {N}$
and
$A_1,\ldots , A_d\in \mathcal {A}$
. Let
$\unicode{x3bb} _i:(0,\infty )\to [0,\infty ) (i=1,\ldots ,d)$
be non-negative functions with
$\unicode{x3bb} _i(t)\to 0 (t\to \infty ,\;i=1,\ldots ,d)$
. Set
Then, there exists a positive constant
$t_0>0$
such that for any
$n,k\in \mathbb {N}_0$
and
$t\geq t_0$
, we have
$$ \begin{align*} R_{n,t}\circ T^k \leq e^{k} R_{n+k,t}. \end{align*} $$
Proof. Take
$t_0>0$
so large that
$\sum _{i=1}^d\unicode{x3bb} _i(t)\leq 1$
for any
$t\geq t_0$
. Since
$S_n^{A_i}\circ T^k\geq S_{n+k}^{A_i}-k$
, we have
$R_{n,t}\circ T^k \leq \exp (\sum _{i=1}^d\unicode{x3bb} _i(t)k) R_{n+k,t}\leq e^k R_{n+k,t}$
for any
$t\geq t_0$
.
5 Proofs of Theorems 3.1 and 3.8
In the following lemma, we give a representation of double Laplace transform of
$Z^Y_n$
in terms of
$Q^Y(s)$
. A similar formula can be found in [Reference Thaler and Zweimüller27, Lemma 7.1].
Lemma 5.1. Let
$s_1>0$
and
$s_2\geq 0$
. Then, we have
$$ \begin{align} &\int_Y \bigg(\sum_{n\geq0} e^{-ns_1}\exp(-s_2 Z^Y_n) \bigg) \bigg(\sum_{n\geq0} e^{-n(s_1+s_2)} \widehat{T}^n 1_{Y_n} \bigg)\,d\mu = \frac{Q^Y (s_1)}{1-e^{-(s_1+s_2)}}. \end{align} $$
Proof. Note that
$Z^Y_n=Z^Y_{n-k}\circ T^k+k$
on
$\{\varphi =k\} (1\leq k \leq n)$
and
$Z^Y_n=0$
on
$\{\varphi>n\}$
, and hence,
$$ \begin{align*} \exp(-s_2 Z^Y_n) = \begin{cases} \exp(-s_2 Z^Y_{n-k}\circ T^k) e^{-k s_2} &\text{on }\{\varphi=k\} (1\leq k \leq n), \\ 1 &\text{on }\{\varphi>n\}. \end{cases} \end{align*} $$
Therefore, for
$n\in \mathbb {N}_0$
,
$$ \begin{align*} \begin{split} &\int_Y (e^{-ns_1}\exp(-s_2 Z^Y_n))\,d\mu \\[5pt] &\quad=\int_Y\bigg(\sum_{k=1}^n e^{-ns_1} \exp(-s_2 Z^Y_{n-k}\circ T^k)e^{-k s_2}1_{Y\cap \{\varphi=k\}}\bigg) \,d\mu +e^{-n s_1}\mu(Y\cap\{\varphi>n\}) \\[5pt] &\quad= \int_Y \sum_{k=1}^n (e^{-(n-k)s_1} \exp(-s_2 Z^Y_{n-k})) (e^{-k(s_1+s_2)}\widehat{T}^k1_{Y\cap\{\varphi=k\}}) \,d\mu \\ &\qquad +e^{-ns_1}\mu(Y\cap\{\varphi>n\}). \end{split} \end{align*} $$
By taking the sum over
$n\in \mathbb {N}_{0}$
, we get
$$ \begin{align*} \begin{split} &\int_Y \bigg( \sum_{n\geq0}e^{-ns_1} \exp(-s_2 Z^Y_n)\bigg)\,d\mu \\ &\quad=\int_Y \bigg(\sum_{n\geq0}e^{-ns_1} \exp(-s_2 Z^Y_n)\bigg) \bigg(\sum_{k\geq1}e^{-k(s_1+s_2)} \widehat{T}^k1_{Y\cap\{\varphi=k\}}\bigg)\,d\mu +Q^Y(s_1), \end{split} \end{align*} $$
and hence,
$$ \begin{align} \int_Y \bigg(\sum_{n\geq0}e^{-ns_1}\exp(-s_2 Z_n)\bigg) \bigg(1_Y-\sum_{k\geq1}e^{-k(s_1+s_2)}\widehat{T}^k1_{Y\cap\{\varphi=k\}}\bigg)\,d\mu = Q^Y(s_1). \end{align} $$
As shown in [Reference Thaler and Zweimüller27, equation (5.3)],
$$ \begin{align} 1_Y-\sum_{k\geq1}e^{-ks}\widehat{T}^k1_{Y\cap\{\varphi=k\}} = (1-e^{-s})\sum_{n\geq0}e^{-ns}\widehat{T}^n 1_{Y_n} \quad\text{a.e.\; }(s>0). \end{align} $$
Lemma 5.2. Assume that conditions (A2) and (A3) of Theorem 3.1 hold. Let
$q>0$
and let
$\unicode{x3bb} :(0,\infty )\to (0,\infty )$
be a positive function with
$\unicode{x3bb} (t)\to 0 (t\to \infty )$
. Then, we have
$$ \begin{align} \int_0^\infty \,e^{-qu}\bigg(\int_Y \,\exp(-\unicode{x3bb}(t)Z^Y_{[ut]})\,d\mu_{H}\bigg)\,du \sim \frac{Q^Y (qt^{-1})}{(q+\unicode{x3bb}(t)t) Q^Y (qt^{-1}+\unicode{x3bb}(t))}, \end{align} $$
as
$t\to \infty $
.
Proof. By substituting
$s_1=qt^{-1}$
and
$s_2=\unicode{x3bb} (t)$
into (5.1), we have
$$ \begin{align} &\int_Y \bigg(\sum_{n\geq0} e^{-nqt^{-1}}\exp(-\unicode{x3bb}(t) Z^Y_n) \bigg) \bigg(\sum_{n\geq0} e^{-n(qt^{-1}+\unicode{x3bb}(t))} \widehat{T}^n 1_{Y_n} \bigg)\,d\mu \nonumber\\[5pt] &\quad=\frac{Q^Y(qt^{-1})}{1-e^{-qt^{-1}-\unicode{x3bb}(t)}} \sim \frac{Q^Y(qt^{-1})}{qt^{-1}+\unicode{x3bb}(t)} \quad (t\to\infty). \end{align} $$
For
$t>0$
, set
$$ \begin{align} H_t= \frac{1}{Q^Y(qt^{-1}+\unicode{x3bb}(t))}\sum_{n\geq0} e^{-n(qt^{-1}+\unicode{x3bb}(t))} \widehat{T}^n 1_{Y_n}. \end{align} $$
Note that condition (A2) implies that the assumptions of Lemma 4.5(2) hold with
$$ \begin{align*} v_n=\widehat{T}^n1_{Y_n}, \quad \int_Y v_n\,d\mu=w_{n+1}^{Y}-w_n^{Y}, \quad \sum_{n\geq0} e^{-ns}\int_Y v_n\,d\mu=Q^Y(s), \end{align*} $$
and
$\unicode{x3bb} (t)$
replaced by
$qt^{-1}+\unicode{x3bb} (t)$
. Hence, there exists
$t_1>0$
such that
$\{H_t\}_{t\geq t_1}$
is uniformly sweeping for
$1_Y$
. In addition, condition (A3) implies that the assumptions in Lemma 4.5(4) are satisfied, and hence,
$H_t\to H$
in
$L^\infty (\mu ) (t\to \infty )$
. Moreover, we can use Lemma 4.6 to take
$t_0\geq t_1$
large enough so that
$\exp (-\unicode{x3bb} (t)Z^Y_n\circ T^k)\leq e^k \exp (-\unicode{x3bb} (t)Z^{Y}_{n+k})$
for any
$t\geq t_0$
and
$n,k\in \mathbb {N}_0$
. Consequently, the assumptions in Lemma 4.4 are fulfilled with
$R_{n,t}=\exp (-\unicode{x3bb} (t)Z^Y_n)$
. Therefore, we can apply Lemmas 4.2 and 4.4 with
$S_n=Z_n^Y$
to get
$$ \begin{align} & \int_Y \bigg(\sum_{n\geq0} e^{-nqt^{-1}}\exp(-\unicode{x3bb}(t) Z^Y_n) \bigg) \bigg(\sum_{n\geq0} e^{-n(qt^{-1}+\unicode{x3bb}(t))} \widehat{T}^n 1_{Y_n} \bigg)\,d\mu \nonumber\\[5pt] &\quad\sim \bigg( \int_Y \sum_{n\geq0} e^{-nqt^{-1}}\exp(-\unicode{x3bb}(t)Z^Y_n) \,d\mu_H\bigg) Q^Y(qt^{-1}+\unicode{x3bb}(t)) \nonumber\\[5pt] &\quad \sim \bigg(t \int_0^\infty e^{-qu}\bigg(\int_Y \exp(-\unicode{x3bb}(t)Z^Y_{[ut]})\,d\mu_H \bigg)\,du \bigg) Q^Y(qt^{-1}+\unicode{x3bb}(t)) \quad(t\to\infty). \end{align} $$
We now prove Theorems 3.1 and 3.8 by using Lemmas 5.1 and 5.2. We imitate the proof of [Reference Kasahara and Yano10, Theorem 2].
Proof of Theorem 3.1
Set
$c(t)=c([t])$
for
$t>0$
. Let
$q,\unicode{x3bb}>0$
be positive constants. By substituting
$\unicode{x3bb} (t)=\unicode{x3bb} /(c(t)t)$
into (5.4), we see that
$$ \begin{align*} \int_0^\infty e^{-qu} \bigg( \int_{Y} \exp\bigg(\!-\frac{\unicode{x3bb} Z^Y_{[ ut]}}{c(t)t} \bigg) \,d\mu_H\bigg)\,du \sim \frac{c(t)Q^Y(qt^{-1})} {\unicode{x3bb} Q^Y(qt^{-1}+\unicode{x3bb} c(t)^{-1}t^{-1})} \quad (t\to\infty). \end{align*} $$
By (2.3), (3.1), and the uniform convergence theorem for regular varying functions [Reference Bingham, Goldie and Teugels6, Theorem 1.5.2], we have
$Q^Y(qt^{-1}+\unicode{x3bb} c(t)^{-1}t^{-1})\sim Q^Y(\unicode{x3bb} c(t)^{-1}t^{-1}) (t\to \infty )$
and
$$ \begin{align} \frac{Q^Y(qt^{-1})}{Q^Y(qt^{-1}+\unicode{x3bb} c(t)^{-1}t^{-1})} &\sim \frac{Q^Y(qt^{-1})}{Q^Y(\unicode{x3bb} c(t)^{-1}t^{-1})} \sim \frac{(qt^{-1})^{-1+\alpha}\ell(q^{-1}t)}{(\unicode{x3bb} c(t)^{-1}t^{-1})^{-1+\alpha}\ell(\unicode{x3bb}^{-1}c(t)t)} \nonumber\\ &\sim \bigg(\frac{q c(t)}{\unicode{x3bb}}\bigg)^{-1+\alpha} \frac{\ell(t)}{\ell(c(t)t)} \quad(t\to\infty). \end{align} $$
Hence,
$$ \begin{align*} & \frac{\ell(c(t)t)}{c(t)^\alpha\ell(t)} \int_0^\infty e^{-qu} \bigg(\int_{Y} \exp\bigg(\!-\frac{\unicode{x3bb} Z^Y_{[ ut]}}{c(t)t} \bigg)\,d\mu_H\bigg)\,du \\ &\quad\to q^{-1+\alpha}\unicode{x3bb}^{-\alpha} = \frac{1}{\Gamma(1-\alpha)} \bigg(\int_0^\infty e^{-qu} u^{-\alpha}\,du\bigg)\unicode{x3bb}^{-\alpha} \quad (t\to\infty). \end{align*} $$
We use Lemma 4.1 to get, for
$0<u<\infty $
,
$$ \begin{align*} &\frac{\ell(c(t)t)}{c(t)^\alpha\ell(t)} \int_{Y} \exp\bigg(\!-\frac{\unicode{x3bb} Z^Y_{[ut]}}{c(t)t} \bigg)\,d\mu_H \\[5pt] \notag &\quad\to \frac{1}{\Gamma(1-\alpha)} u^{-\alpha} \unicode{x3bb}^{-\alpha} = \frac{\sin(\pi\alpha)}{\pi} u^{-\alpha} \int_0^\infty e^{-\unicode{x3bb} s} s^{-1+\alpha}\,ds \quad (t\to\infty). \end{align*} $$
Here, we used Euler’s reflection formula
$\Gamma (\alpha )\Gamma (1-\alpha )=\pi /\sin (\pi \alpha )$
. By the extended continuity theorem for Laplace transforms of locally finite measures [Reference Feller8, Ch. XIII.1, Theorem 2a], for
$0\leq s_0<\infty $
,
$$ \begin{align} &\frac{\ell(c(t)t)}{c(t)^\alpha\ell(t)} \mu_H\bigg(\frac{Z^Y_{[ut]}}{c(t)t} \leq s_0\bigg) \nonumber\\[5pt] &\quad\to \frac{\sin(\pi\alpha)}{\pi} u^{-\alpha} \int_0^{s_0} s^{-1+\alpha}\,ds = \frac{\sin(\pi\alpha)}{\pi\alpha} \bigg(\frac{s_0}{u}\bigg)^\alpha \quad (t\to\infty). \end{align} $$
Therefore, we substitute
$t=n$
and
$u=s_0=1$
into (5.9), and then obtain
$$ \begin{align*} \mu_H\bigg(\frac{Z^Y_n}{n} \leq c(n) \bigg) \sim \frac{\sin(\pi\alpha)}{\pi\alpha}\frac{c(n)^\alpha\ell(n)}{\ell(c(n)n)} \quad(n\to\infty), \end{align*} $$
which is the desired result.
Proof of Theorem 3.8
Set
$c(t)=c([t])$
for
$t>0$
. By Chebyshev’s inequality,
$$ \begin{align} \mu_G \bigg(\frac{Z^Y_{[t]}}{t}\leq c(t)\bigg) \leq e \int_X \exp\bigg(- \frac{Z^Y_{[t]}}{c(t)t}\bigg) \,d\mu_G. \end{align} $$
For each
$t>0$
, the map
$(0,\infty )\ni u\mapsto \int _X \exp (-Z_{[ut]}/(c(t)t))\,d\mu _G\in [0,\infty )$
is non-increasing. Hence, we have
$$ \begin{align} \int_X \exp\bigg(- \frac{Z^Y_{[t]}}{c(t)t}\bigg) \,d\mu_G &\leq \int_0^1 \bigg( \int_X \exp\bigg(- \frac{Z^Y_{[ut]}}{c(t)t}\bigg)\,d\mu_G \bigg)\,du \nonumber\\ &\leq e \int_0^\infty e^{-u} \bigg( \int_X \exp\bigg(- \frac{Z^Y_{[ut]}}{c(t)t}\bigg) \, d\mu_G \bigg) \,du \nonumber \\ &\leq et^{-1} \sum_{n\geq0} e^{-nt^{-1}} \int_X \exp\bigg(- \frac{Z^Y_{n}}{c(t)t}\bigg) G\,d\mu. \end{align} $$
Here, we also used (4.2). Define
$H_t (t>0)$
as in (5.6) with
$q=1$
. As shown in the proof of Lemma 5.2, there exists
$t_1>0$
such that
$\{H_t\}_{t\geq t_1}$
is uniformly sweeping for
$1_Y$
. Since
$1_Y$
is uniformly sweeping for G, so is
$\{H_t\}_{t\geq t_1}$
. Moreover, by Lemma 4.6, we can take
$t_0\geq t_1$
large enough so that
$\exp (-Z^Y_n\circ T^k/(c(t)t))\leq e^k \exp (-Z^{Y}_{n+k}/(c(t)t))$
for any
$t\geq t_0$
and
$n,k\in \mathbb {N}_0$
. Consequently, the assumptions in Lemma 4.3 are fulfilled with
$R_{n,t}=\exp (-Z_n^Y/(c(t)t))$
. Therefore, Lemma 4.3 implies
$$ \begin{align} \sup_{t\geq t_0} \frac{\sum_{n\geq0} e^{-nt^{-1}} \int_X \exp(-Z^Y_n/(c(t)t)) G\,d\mu} {\sum_{n\geq0} e^{-nt^{-1}} \int_{X} \exp(-Z^Y_n/(c(t)t)) H_t\,d\mu} <\infty. \end{align} $$
By substituting
$q=1$
and
$\unicode{x3bb} (t)=c(t)^{-1}t^{-1}$
into (5.5) and making a similar estimate as in (5.8), we see that
$$ \begin{align} &t^{-1}\sum_{n\geq0} e^{-nt^{-1}} \int_X \exp\bigg(- \frac{Z^Y_{n}}{c(t)t}\bigg) H_t\,d\mu \nonumber\\ &\quad= \frac{t^{-1}}{1-\exp(t^{-1}+c(t)^{-1}t^{-1})} \cdot \frac{Q^{Y}(t^{-1})}{Q^Y(t^{-1}+c(t)^{-1}t^{-1})} \sim \frac{ c(t)^\alpha\ell(t)}{\ell(c(t)t)} \quad(t\to\infty). \end{align} $$
Combining (5.10) with (5.11), (5.12), (5.13), we obtain
$$ \begin{align*} \limsup_{t\to\infty} \frac{\ell(c(t)t)}{c(t)^\alpha\ell(t)} \mu_G \bigg(\frac{Z^Y_{[t]}}{t}\leq c(t)\bigg) <\infty, \end{align*} $$
as desired.
6 Proof of Theorem 3.10
Let us represent double Laplace transform of
$S^Y_n$
in terms of
$Q^Y(s)$
. We also refer the reader to [Reference Thaler and Zweimüller27, Lemma 5.1] for a similar formula.
Lemma 6.1. Let
$s_1>0$
and
$s_2\geq 0$
. Then, we have
$$ \begin{align} & (1-e^{-s_2})\int_Y \bigg(\sum_{n\geq0} e^{-ns_1}\exp(-s_2 S_n^Y) \bigg)\,d\mu \nonumber\\ &\qquad + (1-e^{-s_1})e^{-s_2} \int_Y \bigg(\sum_{n\geq0} e^{-ns_1}\exp(-s_2 S_n^Y) \bigg) \bigg(\sum_{n\geq0} e^{-ns_1} \widehat{T}^n 1_{Y_n} \bigg)\,d\mu \nonumber\\ &\quad= Q^Y(s_1). \end{align} $$
Proof. It is easy to see that
$S_n^Y=S_{n-k}^Y\circ T^k+1$
on
$\{\varphi =k\} (1\leq k \leq n)$
and
$S_n^Y=0$
on
$\{\varphi>n\}$
, which implies
$$ \begin{align*} \exp(-s_2 S_n^Y) = \begin{cases} \exp(-s_2 S_{n-k}^Y\circ T^k)e^{-s_2} &\text{on }\{\varphi=k\}, 1\leq k \leq n, \\ 1 &\text{on }\{\varphi>n\}. \end{cases} \end{align*} $$
Thus, for
$n\in \mathbb {N}_0$
,
$$ \begin{align*} &\int_Y (e^{-ns_1}\exp(-s_2 S_n^Y))\,d\mu \\ &\quad= \int_Y \sum_{k=1}^n(e^{-ns_1}\exp(-s_2 S_{n-k}^Y\circ T^k)e^{-s_2} 1_{Y\cap \{\varphi=k\}})\,d\mu +e^{-ns_1}\mu(Y\cap \{\varphi>n\}) \\ &\quad= e^{-s_2}\int_Y \sum_{k=1}^n (e^{-(n-k)s_1}\exp(-\unicode{x3bb}(t)S_{n-k}^Y)) (e^{-ks_2}\widehat{T}^k1_{Y\cap\{\varphi=k\}})\,d\mu \\ &\qquad +e^{-ns_1}\mu(Y\cap\{\varphi>n\}). \end{align*} $$
By taking the sum over
$n\in \mathbb {N}_0$
, we get
$$ \begin{align*} &\int_Y \bigg(\sum_{n\geq0}e^{-ns_1}\exp(-s_2 S_n^Y)\bigg)\,d\mu \\ &\quad= e^{-s_2}\int_Y \bigg(\sum_{n\geq0} e^{-ns_1} \exp(-s_2 S_{n}^Y)\bigg) \bigg(\sum_{k\geq0}e^{-ks_1}\widehat{T}^k1_{Y\cap\{\varphi=k\}}\bigg)\,d\mu +Q^Y(s_1), \end{align*} $$
and hence,
$$ \begin{align} &(1-e^{-s_2})\int_Y \bigg(\sum_{n\geq0}e^{-ns_1}\exp(-s_2 S_n^Y)\bigg)\,d\mu \nonumber\\ &\qquad+ e^{-s_2}\int_Y \bigg(\sum_{n\geq0} e^{-ns_1} \exp(-s_2 S_{n}^Y)\bigg)\bigg(1_Y-\sum_{k\geq0}e^{-ks_1}\widehat{T}^k1_{Y\cap\{\varphi=k\}}\bigg)\,d\mu \nonumber\\ &\quad=Q^Y(s_1). \end{align} $$
Lemma 6.2. Assume that condition (B2) of Theorem 3.10 holds. Let
$q>0$
and let
$\unicode{x3bb} :(0,\infty )\to (0,\infty )$
be a positive function with
$$ \begin{align} \unicode{x3bb}(t)\to0\quad \text{and} \quad \frac{Q^Y(qt^{-1})}{\unicode{x3bb}(t)t}\to 0 \quad (t\to\infty). \end{align} $$
Then, we have
$$ \begin{align} \int_0^\infty e^{-qu} \bigg( \int_Y \exp(-\unicode{x3bb}(t)S_{[ut]}^Y)\,d\mu\bigg) \,du \sim \frac{Q^Y(qt^{-1})}{\unicode{x3bb}(t)t} \quad(t\to\infty). \end{align} $$
Proof. By substituting
$s_1=qt^{-1}$
and
$s_2=\unicode{x3bb} (t)$
into (6.1), we have
$$ \begin{align} & (1-e^{-\unicode{x3bb}(t)})\int_Y \bigg(\sum_{n\geq0} e^{-nqt^{-1}}\exp(-\unicode{x3bb}(t) S_n^Y) \bigg)\,d\mu \nonumber\\ &\qquad + (1-e^{-qt^{-1}})e^{-\unicode{x3bb}(t)} \int_Y \bigg(\sum_{n\geq0} e^{-nqt^{-1}}\exp(-\unicode{x3bb}(t) S_n^Y) \bigg) \bigg(\sum_{n\geq0} e^{-nqt^{-1}} \widehat{T}^n 1_{Y_n} \bigg)\,d\mu \nonumber\\ &\quad= Q^Y(qt^{-1}). \end{align} $$
Let
$l_1(t)$
and
$l_2(t)$
denote the first and second terms of the left-hand side of (6.5), respectively. Let us prove
$l_2(t)$
is negligibly small as
$t\to \infty $
. By condition (B2), the assumptions of Lemma 4.5(3) are satisfied with
$$ \begin{align*} v_n=\widehat{T}^n1_{Y_n}, \quad \sum_{n\geq0} e^{-ns}\int_Y v_n\,d\mu=Q^Y(s), \quad H_t=\frac{1}{Q^Y(qt^{-1})}\sum_{n\geq0} e^{-nqt^{-1}} \widehat{T}^n 1_{Y_n}. \end{align*} $$
Hence, by Lemma 4.5(3), there exists
$C>0$
such that
$$ \begin{align} \bigg\| \sum_{n\geq0} e^{-nqt^{-1}} \widehat{T}^n 1_{Y_n} \bigg\|_{L^\infty(\mu)} \leq C Q^Y(qt^{-1}). \end{align} $$
$$ \begin{align} 0&\leq \frac{l_2(t)} {l_1(t)} \leq \frac{C(1-e^{-qt^{-1}})Q^Y(qt^{-1})} {e^{\unicode{x3bb}(t)}-1} \leq \frac{C qQ^Y(qt^{-1})} {\unicode{x3bb}(t)t} \to0 \quad (t\to\infty). \end{align} $$
However, Lemma 4.2 yields
$$ \begin{align} l_1(t) \sim \unicode{x3bb}(t)t \int_0^\infty e^{-qu} \bigg( \int_Y \exp(-\unicode{x3bb}(t)S_{[ut]}^Y)\,d\mu\bigg)\,du \quad(t\to\infty). \end{align} $$
We now prove Theorem 3.10 by using Lemma 6.2.
Proof of Theorem 3.10
Set
$\widetilde {c}(t)=\widetilde {c}([t])$
for
$t>0$
. Let
$q,\unicode{x3bb}>0$
be positive constants. By substituting
$\unicode{x3bb} (t)=\unicode{x3bb} /(\widetilde {c}(t)a(t))$
into (6.4), we have
$$ \begin{align} \frac{1}{\widetilde c(t)} \int_0^\infty e^{-qu} \bigg( \int_Y \exp\bigg(\!-\frac{\unicode{x3bb} S^{Y}_{[ut]}}{\widetilde{c}(t)a(t)} \bigg)\,d\mu\bigg)\,du \sim \frac{1}{\Gamma(1+\alpha)}\frac{Q^Y(qt^{-1})} {\unicode{x3bb} Q^Y(t^{-1})} \quad(t\to\infty). \end{align} $$
Let
$r(t)$
denote the right-hand side of (6.9). By (2.3),
$$ \begin{align*} r(t) \to \frac{q^{-1+\alpha}\unicode{x3bb}^{-1}}{\Gamma(1+\alpha)} = \frac{\sin(\pi\alpha)}{\pi\alpha} \bigg(\int_0^\infty e^{-qu}u^{-\alpha}\,du\bigg) \bigg(\int_0^\infty e^{-\unicode{x3bb} s}\,ds\bigg) \quad(t\to\infty). \end{align*} $$
Hence, we use Lemma 4.1 to get, for
$0<u<\infty $
,
$$ \begin{align*} \frac{1}{\widetilde{c}(t)} \int_Y \exp\bigg(\!-\frac{\unicode{x3bb} S^{Y}_{[ut]}}{\widetilde{c}(t)a(t)} \bigg)\,d\mu \to \frac{\sin(\pi\alpha)}{\pi\alpha} u^{-\alpha}\int_0^\infty e^{-\unicode{x3bb} s}\, ds \quad (t\to\infty). \end{align*} $$
By the extended continuity theorem for Laplace transforms, for
$0\leq s_0<\infty $
,
$$ \begin{align*} \frac{1}{\widetilde{c}(t)}\mu_{1_Y} \bigg(\frac{S_{[ut]}^Y}{\widetilde{c}(t)a(t)}\leq s_0\bigg) \to \frac{\sin(\pi\alpha)}{\pi\alpha} u^{-\alpha} \int_0^{s_0}ds=\frac{\sin(\pi\alpha)}{\pi\alpha}\frac{s_0}{u^\alpha} \quad (t\to\infty). \end{align*} $$
Substituting
$t=n$
and
$u=s_0=1$
completes the proof.
7 Proofs of Theorems 3.15 and 3.21
We can also represent double Laplace transform of
$S_n^{A_i} (i=1,\ldots ,d)$
in terms of
$Q^{Y,A_i}(s) (i=1,\ldots ,d)$
. We refer the reader to [Reference Thaler and Zweimüller27, Lemma 6.1] and [Reference Sera and Yano18, Proposition 5.1] for similar formulas.
Lemma 7.1. Suppose condition (C1) of Theorem 3.15 is satisfied. Let
$s>0$
and
$s_1,s_2,\ldots ,s_d\geq 0$
. Then, we have
$$ \begin{align} & (1-e^{-s})\int_Y \bigg( \sum_{n\geq0}e^{-ns}\exp \bigg(-\sum_{j=1}^d s_j S_n^{A_j}\bigg) \bigg)\,d\mu \nonumber\\[5pt] &\qquad+\sum_{i=1}^d (e^{s_i}-e^{-s}) \int_Y \bigg( \sum_{n\geq0}e^{-ns}\exp \bigg(-\sum_{j=1}^d s_j S_n^{A_j}\bigg) \bigg) \bigg( \sum_{n\geq1}e^{-n(s+s_i)} \widehat{T}^n 1_{Y_n\cap A_i} \bigg)\,d\mu \nonumber\\[5pt] &\quad= \mu(Y)+ \sum_{i=1}^d Q^{Y,A_i}(s+s_i). \end{align} $$
Proof. Set
$$ \begin{align*} R_{n} =\exp \bigg(-\sum_{i=1}^d s_i S_n^{A_i}\bigg), \quad n\in\mathbb{N}_0. \end{align*} $$
Note that, for
$n\in \mathbb {N}$
,
$$ \begin{align*} R_{n} = \begin{cases} R_{n-1}\circ T &\text{on }\{\varphi=1\}=T^{-1}Y, \\ (R_{n-k}\circ T^{k})e^{-(k-1)s_i} &\text{on }(T^{-1}A_i)\cap \{\varphi=k\} (1\leq i \leq d\ \text{and}\ 2\leq k \leq n), \\ e^{-ns_i} &\text{on }(T^{-1}A_i)\cap \{\varphi> n\} (1\leq i \leq d). \end{cases} \end{align*} $$
Hence,
$\int _Y R_{0}\,d\mu =\mu (Y)$
and, for
$n\in \mathbb {N}$
,
$$ \begin{align*} &\int_Y e^{-ns}R_{n}\,d\mu \\ &\quad= \int_{Y} e^{-ns}R_{n-1}\:\widehat{T}1_{Y\cap T^{-1}Y}\, d\mu \\ &\qquad+ e^{-s} \sum_{i=1}^d \int_Y \sum_{k=2}^{n} (e^{-(n-k)s}R_{n-k})\: ( e^{-(k-1)(s+s_i)} \widehat{T}^{k}1_{Y\cap (T^{-1}A_i)\cap\{\varphi=k\}} ) \,d\mu \\ &\qquad+\sum_{i=1}^d e^{-n(s+s_i)}\mu(Y\cap (T^{-1}A_i)\cap\{\varphi>n\}). \end{align*} $$
By taking the sum over
$n\in \mathbb {N}_0$
, we get
$$ \begin{align*} &\int_Y \bigg(\sum_{n\geq0}e^{-ns}R_{n}\bigg)\,d\mu \\ &\quad= \mu(Y)+e^{-s}\int_Y \bigg(\sum_{n\geq0}e^{-ns}R_{n}\bigg)\widehat{T}1_{Y\cap T^{-1}Y}\,d\mu \\ &\qquad+ e^{-s} \sum_{i=1}^d \int_Y \bigg(\sum_{n\geq0} e^{-ns}R_{n}\bigg)\: \bigg( \sum_{k\geq2} e^{-(k-1)(s+s_i)} \widehat{T}^{k}1_{Y\cap (T^{-1}A_i)\cap\{\varphi=k\}} \bigg)\, d\mu \\ &\qquad+\sum_{i=1}^d Q^{Y,A_i}(s+s_i). \end{align*} $$
By (3.14), we have
$$ \begin{align*} 1_Y&=\widehat{T}1_{T^{-1}Y} =\widehat{T}1_{Y\cap T^{-1}Y} +\sum_{i=1}^d\widehat{T}1_{A_i\cap T^{-1}Y} \\ &= \widehat{T}1_{Y\cap T^{-1}Y} +\sum_{i=1}^d \sum_{k\geq2}\widehat{T}^k 1_{Y\cap (T^{-1}A_i)\cap \{\varphi=k\}}, \end{align*} $$
which implies
$$ \begin{align} &(1-e^{-s})\int_Y \bigg(\sum_{n\geq0}e^{-ns}R_{n}\bigg)\,d\mu \nonumber\\ &\qquad+\sum_{i=1}^d e^{-s} \int_Y \bigg( \sum_{n\geq0}e^{-ns}R_{n}\bigg) \bigg(\sum_{k\geq2}(1-e^{-(k-1)(s+s_i)}) \widehat{T}^k 1_{Y\cap (T^{-1}A_i)\cap \{\varphi=k\}} \bigg)\,d\mu \nonumber\\ &\quad =\mu(Y)+\sum_{i=1}^d Q^{Y,A_i}(s+s_i). \end{align} $$
In addition, we use (3.14) to get, for
$t>0$
,
$$ \begin{align} \sum_{k\geq2}(1-e^{-(k-1)t}) \widehat{T}^k 1_{Y\cap (T^{-1}A_i)\cap \{\varphi=k\}} & = \sum_{k\geq2} \bigg((e^t-1)\sum_{n=1}^{k-1}e^{-nt}\bigg)\widehat{T}^k 1_{Y\cap (T^{-1}A_i)\cap \{\varphi=k\}} \nonumber\\ &= (e^t-1)\sum_{n\geq1} e^{-nt}\sum_{k> n}\widehat{T}^k 1_{Y\cap (T^{-1}A_i)\cap \{\varphi=k\}} \nonumber\\ &= (e^t-1)\sum_{n\geq1} e^{-nt}\:\widehat{T}^{n}1_{Y_n\cap A_i}. \end{align} $$
Lemma 7.2. Assume that conditions (C1), (C3), and (C4) of Theorem 3.15 hold. Let
$q>0$
be a positive constant and let
$\unicode{x3bb} _i:(0,\infty )\to (0,\infty ) (i=1,\ldots ,d-1)$
be positive functions with
as
$t\to \infty $
. Then,
Proof. Set
By substituting
$s=qt^{-1}$
,
$s_i=\unicode{x3bb} _i(t) (i=1,\ldots ,d-1)$
, and
$s_d=0$
into (7.1), we have
Let
$l_1(t),l_2(t)$
, and
$l_3(t)$
denote the first, second, and third terms of the left-hand side of (7.8), respectively, and
$r(t)$
denote the right-hand side of (7.8). Note that
$$ \begin{align} r(t) \sim Q^{Y,A_d}(qt^{-1}) \quad(t\to\infty), \end{align} $$
since
$Q^{Y,A_i}(s)\to \infty (s\to 0+,\; i=1,\ldots ,d)$
and the assumption (7.5).
Let us prove
$l_2(t)$
is the leading term of the left-hand side of (7.8), and
$l_1(t)$
and
$l_3(t)$
are negligible as
$t\to \infty $
. Note that condition (C4) implies that the assumptions in Lemma 4.5(2) are satisfied with
$$ \begin{align} &v_n=\widehat{T}^n 1_{Y_n\cap A_i}, \, \int_Y v_n\,d\mu=w^{Y,A_i}_{n+1}-w^{Y,A_i}_n,\, \sum_{n\geq0}e^{-ns}\int_Y v_n\,d\mu =Q^{Y,A_i}(s), \end{align} $$
Hence, by Lemma 4.5(2), there exists
$t_1>0$
such that
$\{H_t^{(i)}\}_{t\geq t_1,\;i=1,\ldots ,d-1}$
is uniformly sweeping for
$1_Y$
. In addition, we use Lemma 4.7 to take
$t_0\geq t_1$
large enough so that
$R_{n,t}\circ T^k\leq e^k R_{n+k,t}$
for any
$n,k\in \mathbb {N}_0$
and
$t\geq t_0$
. Consequently, the assumptions in Lemma 4.3 are fulfilled with
$H_t=H^{(i)}_t (i=1,\ldots ,d-1)$
and
$G=1_Y$
. Therefore, Lemma 4.3 implies
$$ \begin{align*} C=\limsup_{t\to\infty} \max_{1\leq i\leq d-1} \frac {\int_Y (\sum_{n\geq0}e^{-nqt^{-1}}R_{n,t})\,d\mu} {\int_Y (\sum_{n\geq0}e^{-nqt^{-1}}R_{n,t})H_t^{(i)}\,d\mu} <\infty. \end{align*} $$
Hence,
In addition,
$\{H_t^{(d)}\}_{t>0}$
is
$L^\infty (\mu )$
-bounded, which follows from condition (C3) and Lemma 4.5(3) with
$$ \begin{align*} v_n=\widehat{T}^n 1_{Y_n\cap A_d}, \quad \sum_{n\geq0}e^{-ns}\int_Y v_n\,d\mu = Q^{Y,A_d}(s), \quad H_t=H_t^{(d)}. \end{align*} $$
We use assumption (7.5) to see
Therefore, we get
We now prove Theorems 3.15 and 3.21 by using Lemma 7.2.
Proof of Theorem 3.15
Set
$c(t)=c([t])$
for
$t>0$
. Let
$q,\unicode{x3bb} ,\unicode{x3bb} _1,\ldots ,\unicode{x3bb} _{d-1}>0$
and
$\unicode{x3bb} _i(t)=\unicode{x3bb} \unicode{x3bb} _i/(c(t)t)$
. By (3.1), (3.16), and the uniform convergence theorem for regular varying functions, we see
$Q^{Y,A_i}(qt^{-1}+\unicode{x3bb} _i(t))\sim Q^{Y,A_i}(\unicode{x3bb} _i(t)) (t\to \infty ,\;i=1,\ldots ,d-1)$
. By the Potter bounds for slowly varying functions, we see that
$c(t)^{-1+\alpha }\ell (t)/\ell (c(t)t)\to \infty $
and
$c(t)^{\alpha }\ell (t)/\ell (c(t)t)\to 0 (t\to \infty )$
. Thus, for
$i=1,\ldots ,d-1$
,
and
Therefore, (7.4) and (7.5) are fulfilled. Define
$R_{n,t} (n\in \mathbb {N}_0,\;t>0)$
and
$H_t^{(i)} (t>0, i=1,\ldots ,d-1)$
as in (7.6) and (7.7), respectively. By Lemma 7.2,
Set
$\widetilde {H}_t=\sum _{i=1}^{d-1}\beta _i\unicode{x3bb} _i^\alpha H_t^{(i)} (t>0)$
. Then, we use (7.13) and (7.14) to get
$$ \begin{align} & \frac{\ell(c(t)t)}{c(t)^\alpha \ell(t)} t^{-1}\int_Y \bigg( \sum_{n\geq0} e^{-nqt^{-1}} R_{n,t}\bigg) \widetilde{H}_t\,d\mu \to \beta_d q^{-1+\alpha}\unicode{x3bb}^{-\alpha} \quad (t\to\infty). \end{align} $$
As shown in the proof of Lemma 7.2, there exists
$t_0>0$
such that the assumptions of Lemma 4.3 are satisfied with
$H_t=H_t^{(i)} (i=1,\ldots ,d-1)$
and
$G=1_Y$
, and hence, with
$H_t=\widetilde {H}_t$
and
$G=1_Y$
. Similarly, condition (C5) implies that the assumptions in Lemma 4.5(4) are satisfied with (7.10), (7.11), and
$H=H^{(i)} (i=1,\ldots ,d-1)$
. Therefore, Lemma 4.5(4) implies that
$H_t^{(i)} \to H^{(i)}$
in
$L^\infty (\mu ) (t\to \infty , i=1,\ldots , d-1)$
, and hence,
$\widetilde {H}_t\to \widetilde {H}=\sum _{i=1}^{d-1}\beta _i \unicode{x3bb} _i^\alpha H^{(i)}$
in
$L^\infty (\mu ) (t\to \infty )$
. Consequently, the assumptions of Lemma 4.4 are fulfilled with
$H_t=\widetilde {H}_t$
and
$H=\widetilde {H}$
. Therefore, we use Lemmas 4.2 and 4.4 to get
$$ \begin{align} t^{-1}\int_Y\bigg( \sum_{n\geq0} e^{-nqt^{-1}} R_{n,t}\bigg) \widetilde{H}_t\,d\mu &\sim t^{-1}\int_Y\bigg( \sum_{n\geq0} e^{-nqt^{-1}} R_{n,t}\bigg) \widetilde{H}\,d\mu \nonumber\\ &\sim \int_0^\infty e^{-qu} \bigg( \int_Y R_{[ut],t} \,d\mu_{\widetilde{H}}\bigg) \,du \quad(t\to\infty). \end{align} $$
$$ \begin{align*} \frac{\ell(c(t)t)}{c(t)^\alpha \ell(t)} \int_0^\infty e^{-qu} \bigg( \int_Y R_{[ut],t} \,d\mu_{\widetilde{H}}\bigg) \,du \to \beta_d q^{-1+\alpha}\unicode{x3bb}^{-\alpha} \quad (t\to\infty). \end{align*} $$
Thus, we use similar arguments as in the proof of Theorem 3.1 to obtain
Substituting
$t=n$
and
$s_0=u=1$
completes the proof.
Proof of Theorem 3.21
Set
$c(t)=c([t]) (t>0)$
and
$\unicode{x3bb} _i(t)=\unicode{x3bb} _i/(c(t)t) (t>0,\;i=1,\ldots ,d-1)$
. Define
$R_{n,t} (n\in \mathbb {N}_0,t>0)$
as in (7.6). By Chebyshev’s inequality,
For each
$t>0$
, the map
$(0,\infty )\ni u\mapsto \int _X R_{[ut],t}\,d\mu _G\in [0,\infty )$
is non-increasing. Hence, we have
$$ \begin{align} \int_X R_{[t],t} \,d\mu_G &\leq \int_0^1 \bigg( \int_X R_{[ut],t}\,d\mu_G\bigg)\,du \nonumber\\ &\leq e \int_0^\infty e^{-u} \bigg( \int_X R_{[ut],t}\,d\mu_G\bigg)\,du \nonumber\\ &\leq et^{-1} \sum_{n\geq0} e^{-nt^{-1}} \int_X R_{n,t} G\,d\mu. \end{align} $$
Define
$H_t^{(i)} (t>0,\;i=1,\ldots ,d-1)$
by (7.7) with
$q=1$
, and set
$\widetilde {H}_t=\sum _{i=1}^{d-1}\beta _i\unicode{x3bb} _i^{\alpha } H_t^{(i)}$
. Recall from the proofs of Lemma 7.2 and Theorem 3.15 that there exists
$t_1>0$
such that
$\{\widetilde {H}_t\}_{t\geq t_1}$
is uniformly sweeping for
$1_Y$
, and hence, for G. In addition, by Lemma 4.7, we can take
$t_0\geq t_1$
large enough so that
$R_{n,t}\circ T^k\leq e^k R_{n+k,t}$
for any
$n,k\in \mathbb {N}_0$
and
$t\geq t_0$
. Consequently, the assumptions in Lemma 4.3 are fulfilled with
$H_t=\widetilde {H}_t$
. Therefore, Lemma 4.3 implies
$$ \begin{align} \sup_{t\geq t_0} \frac{\sum_{n\geq0} e^{-nt^{-1}} \int_X R_{n,t}G \,d\mu} {\sum_{n\geq0} e^{-nt^{-1}} \int_X R_{n,t}\widetilde{H}_t \,d\mu} <\infty. \end{align} $$
By substituting
$q=\unicode{x3bb} =1$
into (7.15), we get
$$ \begin{align} &t^{-1} \sum_{n\geq0} e^{-nt^{-1}} \int_X R_{n,t}\widetilde{H}_t \,d\mu \sim \beta_d \frac{c(t)^\alpha\ell(t)}{\ell(c(t)t)} \quad(t\to\infty). \end{align} $$
8 Applications to Thaler’s maps
Our abstract results in §3 are applicable to a variety of classes of ergodic transformations. Indeed, the assumptions of Theorems 3.1, 3.10, and 3.15 are milder than those of [Reference Thaler and Zweimüller27, Theorems 3.3, 3.1, and 3.2], respectively, which are verified for interval maps with indifferent fixed points (see [Reference Thaler and Zweimüller27, §8] and [Reference Sera and Yano18, §2.4]), Markov chains on multiray [Reference Sera and Yano18, §2.5], and random iterations of piecewise linear maps (as summarized in [Reference Hata and Yano9, Theorem 1.1, the subsequent paragraph, and Lemma 3.5] and [Reference Nakamura, Nakano, Toyokawa and Yano13, Theorem 1.2, Remark 1.4, and §4.2]) under suitable settings. The assumptions of Theorems 3.8, 3.10, and 3.21 are also verified for random walks driven by Gibbs–Markov maps, as shown in [Reference Zweimüller33, §7.3]. For simplicity, we are going to focus only on Thaler’s maps with two indifferent fixed points [Reference Thaler25] in this section.
Assumption 8.1. (Thaler’s map)
Suppose that the map
$T:[0,1]\to [0,1]$
satisfies the following conditions.
-
(i) For some
$c\in (0,1)$
, the restrictions
$T:[0,c)\to [0,1)$
and
$T:(c,1]\to (0,1]$
are strictly increasing, onto, and can be extended to
$C^2$
maps
$T_0:[0,c]\to [0,1]$
and
$T_1:[c,1]\to [0,1]$
, respectively. -
(ii)
$T_0'>1$
and
$T_0"> 0$
on
$(0,c]$
,
$T_1'>1$
and
$T^{\prime \prime }_1<0$
on
$[c,1)$
, and
$T'(0)=T'(1)=1$
. -
(iii) For some
$p\in (1,\infty )$
,
$a\in (0,\infty )$
, and some positive, measurable function
$\ell ^*:(0,\infty )\to (0,\infty )$
slowly varying at
$0$
such that
$$ \begin{align*} Tx-x\sim a^{-p}(1-x-T(1-x)) \sim x^{p+1}\ell^*(x) \quad(x\to0+). \end{align*} $$
In the following, we always impose Assumption 8.1. Let us summarize the facts that are shown in [Reference Thaler22, Reference Thaler23, Reference Thaler25, Reference Thaler and Zweimüller27, Reference Zweimüller30, Reference Zweimüller31]. After that, we will explain applications of our abstract results to Thaler’s maps.
Let
$f_i$
denote the inverse function of
$T_i (i=0,1)$
. Then, T admits an invariant density h of the form
$$ \begin{align*} h(x)=h_0(x)\frac{x(1-x)}{(x-f_0(x))(f_1(x)-x)} \quad(x\in(0,1)), \end{align*} $$
where
$h_0$
is continuous and positive on
$[0,1]$
. In addition, h has bounded variation on
$[\varepsilon ,1-\varepsilon ]$
for any
$\varepsilon \in (0,1/2)$
. Define the
$\sigma $
-finite measure
$\mu $
as
$d\mu (x)=h(x)\,dx$
,
$x\in [0,1]$
. Then,
$\mu ([0,\varepsilon ])=\mu ([1-\varepsilon ,1])=\infty $
for any
$\varepsilon \in (0,1)$
, and T is a CEMPT on the
$\sigma $
-finite measure space
$([0,1],\mathcal {B}([0,1]), \mu )$
.
Since
$f_0(x)\sim x (x\to 0+)$
, we use the uniform convergence theorem for slowly varying functions [Reference Bingham, Goldie and Teugels6, Theorem 1.2.1] to get
$\ell ^*(f_0(x))\sim \ell ^*(x) (x\to 0+)$
and
$$ \begin{align} x-f_0(x) &=T_0(f_0(x))-f_0(x) \nonumber\\ &\sim f_0(x)^{p+1}\ell^*(f_0(x)) \sim x^{p+1}\ell^*(x) \quad (x\to0+). \end{align} $$
Similarly, it is easily seen that
$1-f_1(1-x)\sim x (x\to 0+)$
and
$$ \begin{align} f_1(1-x)-(1-x) &= 1-(1-f_1(1-x)) -T_1(1-(1-f_1(1-x))) \nonumber\\ &\sim a^p x^{p+1}\ell^*(x) \quad (x\to 0+). \end{align} $$
Let
$\gamma \in (0,c)$
be a
$2$
-periodic point of T. Then,
$T\gamma \in (c,1)$
. Take
$c_0\in (0,\gamma ]$
and
$c_1\in [T\gamma ,1)$
arbitrarily, and set
$$ \begin{align} A_0=[0, c_0),\quad Y=[c_0, c_1],\quad A_1=(c_1, 1]. \end{align} $$
Then,
$\mu (Y)\in (0,\infty )$
,
$\mu (A_i)=\infty (i=0,1)$
, and Y dynamically separates
$A_0$
and
$A_1$
.
Lemma 8.2. For
$i=0,1$
, there exists a
$\mu $
-probability density function
$H^{(i)}$
such that
$H^{(i)}$
is positive, continuous, supported, and has bounded variation on
$(TA_i)\setminus A_i$
, and satisfies
$$ \begin{align*} \lim_{n\to\infty} \frac{\widehat{T}^n1_{Y\cap (T^{-1}A_i)\cap \{\varphi=n\} }} { \mu(Y\cap (T^{-1}A_i)\cap \{\varphi=n\})} = H^{(i)} \quad \text{in }L^\infty(\mu)\quad(i=0,1). \end{align*} $$
In addition,
$$ \begin{align} &\mu(Y\cap (T^{-1}A_i)\cap \{\varphi=n\}) = \int_{(TA_i)\setminus A_i} \widehat{T}^n 1_{Y\cap (T^{-1}A_i)\cap\{\varphi=n\}}\,d\mu \nonumber\\ &\sim \begin{cases} h(c)f_1'(0)(f_0^n(1)-f_0^{n+1}(1)) &(n\to\infty, \,i=0), \\ h(c)f_0'(1)(f_1^{n+1}(0)-f_1^n(0))\quad &(n\to\infty,\,i=1). \end{cases} \end{align} $$
Remark 8.3. By Lemmas 3.14 and 8.2, we have
$$ \begin{align} \lim_{n\to\infty}\bigg(\frac{1}{w_n^{Y,A_i}} \sum_{k=0}^{n-1} \widehat{T}^k 1_{Y_k\cap A_i}\bigg) = H^{(i)} \quad \text{in }L^\infty(\mu) \quad(i=0,1). \end{align} $$
Proof of Lemma 8.2
We proceed as in the proofs of [Reference Thaler25, Lemma 3] and [Reference Thaler and Zweimüller27, Theorem 8.1]. As in the calculations in [Reference Thaler25, p. 1301] and [Reference Thaler and Zweimüller27, p. 46], for
$n\geq 2$
,
$$ \begin{align}\hspace{-9pt} \widehat{T}^n1_{Y\cap (T^{-1}A_0)\cap \{\varphi=n\} } &= 1_{(TA_0)\setminus A_0} (h\circ f_1\circ f_0^{n-1})\cdot (f_1'\circ f_0^{n-1})\cdot (f_0^{n-1})'/h \quad\text{a.e.,} \end{align} $$
$$ \begin{align}\hspace{-10pt} \widehat{T}^n1_{Y\cap (T^{-1}A_1)\cap \{\varphi=n\} } &= 1_{(TA_1)\setminus A_1} (h\circ f_0\circ f_1^{n-1})\cdot (f_0'\circ f_1^{n-1})\cdot (f_1^{n-1})'/h \quad\text{a.e.} \end{align} $$
It is easily seen that
$f_0^{n-1}\to 0$
and
$f_1^{n-1}\to 1 (n\to \infty )$
uniformly on
$[0,1]$
. By applying [Reference Thaler25, Lemma 2] to
$f(x)=f_0(x)$
and
$f(x)=1-f_1(1-x)$
, respectively, we see that there exist continuous functions
$g_0:(0,1]\to (0,\infty )$
and
$g_1:[0,1)\to (0,\infty )$
such that
$$ \begin{align} \frac{(f_0^n)'(x)}{f^{n}_0(1)-f^{n+1}_0(1)} \to g_0(x) \quad (n\to\infty),\, \text{uniformly on compact subsets on (0,1]}, \end{align} $$
$$ \begin{align} \frac{(f_1^n)'(x)}{f^{n+1}_1(0)-f^{n}_1(0)} \to g_1(x) \quad (n\to\infty),\, \text{uniformly on compact subsets on [0,1)}, \end{align} $$
and
$$ \begin{align} \int_{f_0(x)}^x g_0(y)\,dy= \int_{x}^{f_1(x)}g_1(y)\,dy= 1 \quad \text{for any }x\in(0,1). \end{align} $$
It follows from the concavity of
$f_0^n$
and (8.8) that
$g_0$
is non-increasing. Similarly,
$g_1$
is non-decreasing, which follows from the convexity of
$f_1^n$
and (8.9). Set
$$ \begin{align*} H^{(i)}=1_{(TA_i)\setminus A_i}g_i/h \quad (i=0,1), \end{align*} $$
which is positive, continuous, supported, and has bounded variation on
$(TA_i)\setminus A_i (i=0,1)$
. By (8.6), (8.7) (8.8), and (8.9),
$$ \begin{align} \lim_{n\to\infty} \frac{\widehat{T}^n1_{Y\cap (T^{-1}A_0)\cap \{\varphi=n\} }} { h(c)f_1'(0)(f_0^n(1)-f_0^{n+1}(1))} = H^{(0)} \quad \text{in }L^\infty(\mu), \end{align} $$
$$ \begin{align} \lim_{n\to\infty} \frac{\widehat{T}^n1_{Y\cap (T^{-1}A_1)\cap \{\varphi=n\} }} { h(c)f_0'(1)(f_1^{n+1}(0)-f_1^n(0))} = H^{(1)} \quad \text{in }L^\infty(\mu). \end{align} $$
By using
$d\mu (x)=h(x)\,dx$
and (8.10), we see that
$$ \begin{align} &\int_{(TA_0)\setminus A_0}H^{(0)}\,d\mu = \int_{c_0}^{T(c_0)}g_0(y)\,dy=1, \end{align} $$
$$ \begin{align} &\int_{(TA_1)\setminus A_1} H^{(1)}\,d\mu = \int_{T(c_1)}^{c_1}g_1(y)\,dy=1. \end{align} $$
Hence,
$H^{(0)}$
and
$H^{(1)}$
are
$\mu $
-probability density functions. We use (8.11), (8.12), (8.13), and (8.14) to obtain the desired result.
Lemma 8.4. There exists a function
$\ell _0(x)$
slowly varying at
$\infty $
such that
$$ \begin{align} &f_0^{n}(1)-f_0^{n+1}(1) \sim p^{-1}n^{-1-(1/p)}\ell_0(n)\quad(n\to \infty), \end{align} $$
$$ \begin{align} &f_1^{n+1}(0)-f_1^{n}(0) \sim (ap)^{-1}n^{-1-(1/p)}\ell_0(n)\quad(n\to \infty). \end{align} $$
Proof. We follow the arguments of [Reference Thaler25, Lemma 5] and [Reference Zweimüller32, Remark 1]. Let
$$ \begin{align} u_0(x)=\int_x^1 \frac{dy}{y-f_0(y)},\quad u_1(x) = \int_x^1 \frac{dy}{f_1(1-y)-(1-y)} \quad(x\in(0,1]). \end{align} $$
We use (8.1), (8.2), and Karamata’s theorem [Reference Bingham, Goldie and Teugels6, Theorem 1.5.11] to see
$$ \begin{align} u_0(x) \sim \frac{1}{px^{p}\ell^*(x)},\quad u_1(x) \sim \frac{1}{p a^px^{p}\ell^*(x)} \quad(x\to0+). \end{align} $$
By (8.18), the map
$x\mapsto u_0(x^{-1})$
is regularly varying at
$\infty $
of index p, and therefore, [Reference Bingham, Goldie and Teugels6, Theorem 1.5.12] implies that its inverse function
$1/u_0^{-1}(x)$
is regularly varying at
$\infty $
of index
$1/p$
. Hence, there exists a function
$\ell _0(x)$
slowly varying at
$\infty $
such that
$$ \begin{align*} u_0^{-1}(x)\sim x^{-1/p}\ell_0(x) \quad(x\to\infty). \end{align*} $$
Since
$u_1(x)\sim a^{-p}u_0(x) (x\to \infty )$
, we have
$$ \begin{align*} u_1^{-1}(x)\sim (a^{-p}u_0)^{-1}(x) = u_0^{-1}(a^p x) \sim a^{-1}x^{-1/p}\ell_0(x) \quad(x\to \infty). \end{align*} $$
Using [Reference Thaler23, Lemma 2] with
$f(x)=f_0(x)$
,
$a_n= f^n_0(1)$
(respectively
$f(x)=1-f_1(1-x)$
and
$a_n=1-f^n_1(0)$
), and
$g(x)\equiv 1$
, we see
$u_0(f^n_0(1))\sim n (n\to \infty )$
(respectively
$u_1(1-f^n_1(0))\sim n (n\to \infty )$
). Hence, it follows from the uniform convergence theorem for regular varying functions [Reference Bingham, Goldie and Teugels6, Theorem 1.5.2] that
$$ \begin{align} f_0^n(1)&=u^{-1}_0 (u_0(f_0^n(1))) \sim u_0^{-1}(n) \sim n^{-1/p}\ell_0(n) \quad(n\to\infty), \end{align} $$
$$ \begin{align} 1-f_1^n(0) &= u_1^{-1}(u_1(1-f_1^n(0))) \sim u_1^{-1}(n) \sim a^{-1}n^{-1/p}\ell_0(n) \quad(n\to \infty).\end{align} $$
Note that
$$ \begin{align*} f_0^n(1)=\sum_{k\geq n}(f_0^k(1)-f_0^{k+1}(1)), \quad 1-f_1^n(0)=\sum_{k\geq n}(f_1^{k+1}(0)-f_1^{k}(0)). \end{align*} $$
In addition,
$(f_0^{n}(1)-f_0^{n+1}(1))_{n\geq 0}$
and
$(f_1^{n+1}(0)-f_1^{n}(0))_{n\geq 0}$
are decreasing sequences, since
$f_0$
and
$f_1$
are
$C^2$
and
$0<f_0'(x), f_1'(y)<1 (x\in (0,1], y\in [0,1))$
. Hence, the desired result follows from (8.19), (8.20), and the monotone density theorem [Reference Bingham, Goldie and Teugels6, Theorem 1.7.2].
Lemma 8.5. Define
$\alpha , \beta _0,\beta _1\in (0,1)$
by
$$ \begin{align*} \alpha=\frac{1}{p},\quad \beta_0 =\frac{f^{\prime}_1(0)}{f^{\prime}_1(0)+f_0'(1)a^{-1}} = \frac{T'(c-)}{T'(c-)+T'(c+)a^{-1}},\quad \beta_1 = 1-\beta_0. \end{align*} $$
Set
$$ \begin{align*} \ell(x)=h(c)(f_1'(0)+f_0'(1)a^{-1})(1-p^{-1})^{-1}\ell_0(x), \end{align*} $$
which is slowly varying at
$\infty $
. Then,
$$ \begin{align} w_n^{Y,A_i}\sim \beta_i n^{1-\alpha}\ell(n) \quad(n\to\infty,\;i=0,1). \end{align} $$
Remark 8.6. For example, if
$\ell ^*(x)\sim C^* (x\to 0+)$
for some constant
$C^*>0$
in Assumption 8.1(iii), then
$\ell (x)\sim C (x\to \infty )$
for some constant
$C>0$
.
Proof of Lemma 8.5
Combining (8.4) with (8.15) and (8.16), we have
$$ \begin{align} &\mu(Y\cap (T^{-1}A_i)\cap\{\varphi=n\}) \nonumber\\ &\quad\sim \begin{cases} h(c)f_1'(0)p^{-1}n^{-1-(1/p)}\ell_0(n) &(n\to\infty,\,i=0) \\ h(c)f_0'(1)(ap)^{-1}n^{-1-(1/p)}\ell_0(n) &(n\to\infty,\,i=1) \end{cases} \nonumber\\ &\quad\sim \alpha(1-\alpha)\beta_i n^{-1-\alpha}\ell(n) \qquad\ \ \ \quad\qquad(n\to\infty,\,i=0,1). \end{align} $$
We use (3.15), (8.22), and apply Karamata’s theorem [Reference Bingham, Goldie and Teugels6, Theorem 1.5.11] twice to obtain
$$ \begin{align*} w^{Y,A_i}_n = \sum_{k=1}^{n-1} \mu(Y\cap (T^{-1}A_i)\cap \{\varphi>k\}) \sim \beta_i n^{1-\alpha}\ell(n) \quad(n\to\infty,\,i=0,1), \end{align*} $$
as desired.
$$ \begin{align*} \lim_{n\to\infty}\bigg(\frac{1}{w_n^{Y}} \sum_{k=0}^{n-1} \widehat{T}^k 1_{Y_k}\bigg) = \beta_0H^{(0)} +\beta_1H^{(1)} =:H \quad \text{in }L^\infty(\mu), \end{align*} $$
and
$$ \begin{align*} w_n^{Y}\sim w_n^{Y,A_0}+w_n^{Y,A_1}\sim n^{1-\alpha}\ell(n) \quad(n\to\infty). \end{align*} $$
Moreover, if
$G:[0,1]\to [0,\infty )$
is Riemann integrable on
$[0,1]$
with
$\int _0^1 G(x)\,dx>0$
, then G is uniformly sweeping for
$1_{[\varepsilon , 1-\varepsilon ]}$
for any
$\varepsilon \in (0,1/2)$
, which follows from [Reference Thaler and Zweimüller27, Theorem 8.1]. Therefore,
$H, H^{(0)}, H^{(1)}$
are uniformly sweeping for
$1_{[\varepsilon , 1-\varepsilon ]}$
and hence, for
$1_Y$
. So we use our main results in §3 to obtain the following theorems.
Theorem 8.7. Let
$\{c(n)\}_{n\geq 0}$
and
$\{\widetilde {c}(n)\}_{n\geq 0}$
be positive sequences satisfying (3.1) and (3.9), respectively. Then, we have (3.2), (3.10), and
$$ \begin{align*} \mu_{H^{(i)}}\bigg(\frac{S_n^{A_i}}{n}\leq c(n)\bigg) \sim \frac{1-\beta_i}{\beta_i} \frac{\sin(\pi \alpha)}{\pi\alpha} \frac{c(n)^\alpha\ell(n)}{\ell(c(n)n)} \quad (n\to\infty,\;i=0,1). \end{align*} $$
Theorem 8.8. Assume
$G\in \{u\in L^1(\mu ):u\geq 0\}$
admits a version that is Riemann integrable on
$[0,1]$
with
$\int _0^1 G(x)\,dx>0$
. Then, there exists some constant
$C_1\in (0,\infty )$
such that, for any positive sequences
$\{c(n)\}_{n\geq 0}$
and
$\{\widetilde {c}(n)\}_{n\geq 0}$
satisfying (3.1) and (3.9), we have (3.6), (3.12), and
$$ \begin{align*} C_1\leq \liminf_{n\to\infty} \frac{\ell(c(n)n)}{c(n)^\alpha\ell(n)} \mu_G\bigg(\frac{S_n^{A_i}}{n}\leq c(n)\bigg) \quad(i=0,1). \end{align*} $$
Theorem 8.9. Assume
$G\in \{u\in L^\infty (\mu ):u\geq 0 \}$
is supported on
$[\varepsilon , 1-\varepsilon ]$
for some
$\varepsilon \in (0,1/2)$
. Then, there exists some constant
$C_2\in (0,\infty )$
such that, for any positive sequences
$\{c(n)\}_{n\geq 0}$
and
$\{\widetilde {c}(n)\}_{n\geq 0}$
satisfying (3.1) and (3.9), we have (3.7), (3.13), and
$$ \begin{align*} \limsup_{n\to\infty} \frac{\ell(c(n)n)}{c(n)^\alpha\ell(n)} \mu_G\bigg(\frac{S_n^{A_i}}{n}\leq c(n)\bigg) \leq C_2 \quad(i=0,1). \end{align*} $$
Remark 8.10. Let
$\nu $
be a probability measure on
$[0,1]$
that is supported on
$[\varepsilon , 1-\varepsilon ]$
for some
$\varepsilon \in (0,1/2)$
and admits a Riemann integrable density u with respect to the Lebesgue measure. Then,
$G=u/h$
is also supported on
$[\varepsilon , 1-\varepsilon ]$
and Riemann integrable, and hence, Theorems 8.8 and 8.9 can be applied to
$\nu =\mu _{G}$
.
Remark 8.11. Let
$0<\alpha <1$
and let
$\ell _i:(0,\infty )\to (0,\infty ) (i=0,1)$
be slowly varying at
$\infty $
. As shown below, there exists
$T:[0,1]\to [0,1]$
satisfying conditions (i) and (ii) of Assumption 8.1 with
$c=1/2$
and
$$ \begin{align} w^{Y, A_i}_n\sim d_i n^{1-\alpha}\ell_i(n) \quad(n\to\infty, \;i=0,1) \end{align} $$
for some constant
$d_i>0$
, where Y and
$A_i$
are chosen as in (8.3). Let us construct such a map T. Let
$\phi _i(x)= x^{-\alpha }\ell _i(x)$
. We may assume that
$\phi _i(x)$
is bounded below on
$(0,R)$
for any
$R>0$
. Applying [Reference Bingham, Goldie and Teugels6, Theorem 1.5.12] to
$f_i(x)=1/\phi _i(x)$
, we can see that the right-continuous inverse
$f_i^{-1}(y):=\sup \{y\in (0,\infty ):f_i(y)>x\}$
is a regularly varying function at
$\infty $
of index
$1/\alpha $
satisfying
$f_i(f_i^{-1}(x))\sim f_i^{-1}(f_i(x))\sim x (x\to \infty )$
. By [Reference Bingham, Goldie and Teugels6, Theorem 1.8.2], we can take a
$C^\infty $
function
$\psi _i:(0,\infty )\to (0,\infty )$
such that
$\psi _i(x)\sim f_i^{-1}(x^{-1}) (x\to 0+)$
. Then,
$\psi _i(x)$
is a regularly varying function at
$0$
of index
$-1/\alpha $
satisfying
$\phi _i(\psi _i(x))\sim x (x\to 0+)$
and
$\psi _i(\phi _i(x))\sim x (x\to \infty )$
. Set
$$ \begin{align*} \Psi_i(x)=\int_0^x \bigg(\int_0^y \frac{dt}{t\psi_i(t)}\bigg)\,dy, \quad x\geq0. \end{align*} $$
Karamata’s theorem [Reference Bingham, Goldie and Teugels6, Theorem 1.5.11] implies that
$\Psi _i(x)\sim \alpha ^2(1+\alpha )^{-1}x/\psi _i(x) (x\to 0+)$
. Take a constant
$b_i>0$
so that
$\Psi _i(b_i/2)=1/2$
. We now define
$T:[0,1]\to [0,1]$
by
$$ \begin{align*} Tx = \begin{cases} x+\Psi_0(b_0 x), &x\in[0,1/2], \\ x-\Psi_1(b_1(1-x)), &x\in(1/2,1]. \end{cases} \end{align*} $$
It is easily seen that T satisfies conditions (i) and (ii) of Assumption 8.1 with
$c=1/2$
. In addition,
$$ \begin{align*} Tx-x\sim \frac{\alpha^2 b_0^{1+1/\alpha}}{1+\alpha}\frac{x}{\psi_0(x)}, \quad (1-x)-T(1-x)\sim \frac{\alpha^2 b_1^{1+1/\alpha}}{1+\alpha}\frac{x}{\psi_1(x)} \quad(x\to0+). \end{align*} $$
Define
$u_0(x)$
and
$u_1(x)$
by (8.17). Then,
$$ \begin{align*} u_i(x)\sim \frac{1+\alpha}{\alpha b_i^{1+1/\alpha}}\psi_i(x) \quad (x\to0+,\;i=0,1), \end{align*} $$
and hence,
$$ \begin{align*} u_i^{-1}(x) \sim \bigg(\frac{1+\alpha}{\alpha b_i^{1+1/\alpha}}\bigg)^{\alpha} \phi_i(x)\quad (x\to\infty,\;i=0,1). \end{align*} $$
Therefore, as in (8.21), we obtain (8.23) for some
$d_i>0$
.
Remark 8.12. Following [Reference Zweimüller33, Example 7.1], let us construct
$T:[0,1]\to [0,1]$
and
$Y\subset [0,1]$
satisfying conditions (A1) and (A2) of Theorem 3.1 and condition (B2) of Theorem 3.10, but violating condition (A3) of Theorem 3.1. We can apply Theorems 3.8 and 3.10, but cannot apply Theorem 3.1 to such T and Y. Let
$\ell _0(x)$
be a slowly varying function at
$\infty $
satisfying
$$ \begin{align} \liminf_{x\to\infty}\ell_0(x)=0, \quad \limsup_{x\to\infty}\ell_0(x)=\infty. \end{align} $$
An example of such a function can be found in [Reference Bingham, Goldie and Teugels6, §1.3.3]. Let
$\ell _1(x)\equiv 1$
. By Remark 8.11, there exists a map
$T:[0,1]\to [0,1]$
satisfying conditions (i) and (ii) of Assumption 8.1 with
$c=1/2$
and (8.23) for some constant
$d_i>0$
, where Y and
$A_i$
are chosen as in (8.3) with
$Tc_0<Tc_1$
. Then,
$(TA_0)\cap (TA_1)=\emptyset $
. In addition,
$$ \begin{align*} w_n^Y\sim w_n^{Y,A_0}+w_n^{Y,A_1}\sim n^{1-\alpha}(d_0\ell_0(n)+d_1) \quad (n\to\infty), \end{align*} $$
and
$d_0\ell _0(x)+d_1$
is slowly varying at
$\infty $
, since, for any
$\unicode{x3bb}>0$
,
$$ \begin{align*} \bigg|\frac{d_0\ell_0(\unicode{x3bb} x)+d_1}{d_0\ell_0(x)+d_1}-1\bigg| = \bigg|\frac{d_0(\ell_0(\unicode{x3bb} x)-\ell_0(x))}{d_0\ell_0(x)+d_1}\bigg| \leq \bigg|\frac{\ell_0(\unicode{x3bb} x)}{\ell_0(x)}-1\bigg|\to0 \quad(x\to\infty). \end{align*} $$
Therefore, condition (A1) is verified. Moreover, there exist
$\mu $
-probability density functions
$H^{(0)}, H^{(1)}:[0,1]\to [0,\infty )$
such that
$H^{(i)}$
is supported and has bounded variation on
$(TA_i)\setminus A_i\subset Y (i=0,1)$
, and (8.5) holds. Note that condition (A3) does not hold because
$H^{(0)}\neq H^{(1)}$
and (8.24). By [Reference Thaler and Zweimüller27, Theorem 8.1],
$H_0$
and
$H_1$
are uniformly sweeping for
$1_Y$
. Hence, there exists
$N\in \mathbb {N}$
such that
$$ \begin{align*} \bigg\{ \frac{1}{w_n^{Y,A_i}} \sum_{k=0}^{n-1}\widehat{T}^k 1_{Y_k\cap A_i} \bigg\}_{n\geq N;\,i=0,1} \end{align*} $$
is
$L^\infty (\mu )$
-bounded and uniformly sweeping for
$1_Y$
. Therefore, conditions (A2) and (B2) are verified, as desired.
Acknowledgements
I am grateful to Professors Kouji Yano and Yuko Yano for valuable discussions, to Professor Alain Rouault for bringing the paper [Reference Rouault, Yor and Zani16] to my attention, and to the referee for careful reading and valuable suggestions. This research was partially supported by JSPS KAKENHI Grant Numbers JP23K19010 and JP24K16948, and by the Research Institute for Mathematical Sciences, an International Joint Usage/Research Center located in Kyoto University.
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