Darling–Kac Theorem for Renewal Shifts in the Absence of Regular Variation

Abstract

Regular variation is an essential condition for the existence of a Darling–Kac law. We weaken this condition assuming that the renewal distribution belongs to the domain of geometric partial attraction of a semistable law. In the simple setting of one-sided null recurrent renewal chains, we derive a Darling–Kac limit theorem along subsequences. Also in this context, we determine the asymptotic behaviour of the renewal function and obtain a Karamata theorem for positive operators. We provide several examples of dynamical systems to which these results apply.

Appendix

1.1 On the Discrete Form of (10)

Let us assume that the discrete version of (10) holds, i.e.

$$\begin{aligned} {\overline{F}}(n) = n^{-\alpha } {\ell (n)} [ M(\delta (n)) + h(n) ], \end{aligned}$$

where $\ell : {{\mathbb {N}}}\rightarrow (0,\infty )$ is a slowly varying sequence, and $h: {{\mathbb {N}}}\rightarrow {{\mathbb {R}}}$ is right-continuous error function such that $\lim _{n \rightarrow \infty } h( \lfloor A_{k_n} x \rfloor ) = 0$, whenever x is a continuity point of M. It is possible to extend the functions $\ell$ and h (still denoted by $\ell$ and h), such that (10) holds. Indeed, let

$$\begin{aligned} \ell (x) = \lfloor x \rfloor ^{-\alpha }{\ell (\lfloor x \rfloor ) x^\alpha }, \quad h(x) = h(\lfloor x \rfloor ) + M(\delta (\lfloor x \rfloor )) - M(\delta (x)). \end{aligned}$$

Then, ${\overline{F}}(x) = x^{-\alpha } {\ell (x)} [ M(\delta (x)) + h(x) ] = {\overline{F}}(\lfloor x \rfloor )$. Clearly, $\ell$ is a slowly varying function, so we only have to show that h satisfies the conditions after (10). Let x be a continuity point of M, and assume that $x \in (1, c^{1/\alpha })$. The general case follows similarly. By the definition

$$\begin{aligned} h(A_{k_n} x ) = h( \lfloor A_{k_n} x \rfloor ) + [M(\delta (\lfloor A_{k_n} x \rfloor )) - M(\delta ( A_{k_n} x))], \end{aligned}$$

so according to assumption on h, it is enough to show that the term in the square brackets tends to 0. As $A_{k_{n+1}} / A_{k_n} \rightarrow c^{1/\alpha }$, we have for n large enough, $\delta ( A_{k_n} x ) = x$, and $\delta ( \lfloor A_{k_n} x \rfloor ) = \lfloor A_{k_n} x \rfloor / A_{k_n} \rightarrow x$. Since x is a continuity point of M, the statement follows.

1.2 Proof of Lemma 3

Introduce the notation

$$\begin{aligned} \begin{aligned} \nu _\lambda ( x)&= 1 - \frac{R_\lambda (x)}{R_\lambda (1)} = 1- x^{-\alpha } \frac{M(x \lambda ^{1/\alpha })}{M(\lambda ^{1/\alpha })}, \quad x \ge 1. \end{aligned} \end{aligned}$$

Then, $\nu _\lambda$ is a distribution function. Consider the decomposition $G_\lambda (x) = G_{\lambda , 1}(x) * G_{\lambda , 2}(x)$, where

$$\begin{aligned} G_{\lambda , 1}(x) = \mathrm{e}^{-s} \sum _{n=0}^\infty \frac{s^n}{n!} \nu _\lambda ^{*n}(x), \end{aligned}$$

with $s= - R_\lambda (1)$, where $*$ stands for convolution, and $*n$ for nth convolution power. Simply

$$\begin{aligned} \int _0^\infty \mathrm{e}^{-u x} \mathrm {d}G_{\lambda , 1}(x) = \exp \left\{ - \int _1^\infty \left( 1 - \mathrm{e}^{-u y} \right) \mathrm {d}R_\lambda (y) \right\} . \end{aligned}$$

Since ${\overline{G}}_{\lambda , 2} (x) = o(\mathrm{e}^{-x})$ holds uniformly in $\lambda$ as $x \rightarrow \infty$, we have that ${\overline{G}}_{\lambda } (x) \sim {\overline{G}}_{\lambda , 1} (x)$ uniformly in $\lambda \in [c^{-1},1]$. Therefore, to prove the statement we have to show that ${\overline{G}}_{\lambda , 1} (x) \sim x^{-\alpha } {M ( x \lambda ^{1/\alpha } )}$ holds uniformly in $\lambda \in [c^{-1},1]$. From the proof of implication (ii) $\Rightarrow$ (iii) of Theorem 3 in [11], we see that it is enough to show that subexponential property and the Kesten bounds hold uniformly in $\lambda \in [c^{-1},1]$, i.e. with ${\overline{\nu }}_\lambda (x) = 1 - \nu _\lambda (x)$

$$\begin{aligned} \lim _{x \rightarrow \infty } \sup _{\lambda \in [c^{-1},1]} \left| \frac{\overline{\nu ^{*n}_\lambda }(x)}{{\overline{\nu }}_\lambda (x)} - n \right| =0, \end{aligned}$$

(51)

and for any $\varepsilon > 0$, there exists K, such that for all $n \in {{\mathbb {N}}}$ and $\lambda \in [c^{-1},1]$

$$\begin{aligned} \overline{\nu ^{*n}_{\lambda }} (x) \le K ( 1 + \varepsilon )^n {\overline{\nu }}_\lambda (x). \end{aligned}$$

(52)

According to Theorem 3.35 and Theorem 3.39 (with $\tau \equiv n$) by Foss et al. [13] both (51) and (52) hold if

$$\begin{aligned} \lim _{x \rightarrow \infty } \sup _{\lambda \in [c^{-1},1]} \left| \frac{\overline{ \nu _\lambda * \nu _\lambda } (x)}{{\overline{\nu }}_\lambda (x)} - 2 \right| = 0. \end{aligned}$$

(53)

Now we prove (53). Write

$$\begin{aligned} \frac{\overline{\nu _\lambda * \nu _\lambda }(x)}{{\overline{\nu }}_\lambda (x)} = \int _1^{x-1} \frac{{\overline{\nu }}_\lambda (x-y)}{{\overline{\nu }}_\lambda (x)} \mathrm {d}\nu _\lambda (y) + \frac{{\overline{\nu }}_\lambda (x-1)}{{\overline{\nu }}_\lambda (x)}. \end{aligned}$$

(54)

By the logarithmic periodicity of M the second term can be written as

$$\begin{aligned} \frac{{\overline{\nu }}_\lambda (x-1)}{{\overline{\nu }}_\lambda (x)} = \left( \frac{x}{x-1} \right) ^\alpha \frac{M \left( c^{\alpha ^{-1} \{ \log _c x^\alpha \}} \lambda ^{1/\alpha } ( 1 - x^{-1}) \right) }{M \left( c^{\alpha ^{-1} \{ \log _c x^\alpha \}} \lambda ^{1/\alpha } \right) }, \end{aligned}$$

which goes to 1 uniformly in $\lambda$ due to the continuity of M (a continuous function is uniformly continuous on compacts). To handle the first term in (54), choose $\delta > 0$ arbitrarily small, and K so large that $\sup _{\lambda \in [c^{-1},1]} {\overline{\nu }}_\lambda (K) < \delta$. As before, uniformly in $\lambda \in [c^{-1},1]$,

$$\begin{aligned} \int _1^K \left| \frac{{\overline{\nu }}_\lambda (x-y)}{{\overline{\nu }}_\lambda (x)} - 1 \right| \mathrm {d}\nu _\lambda (y) \rightarrow 0. \end{aligned}$$

We show that the integral on $[K, \infty )$ is small. Put $C = \sup _{y} M(y) / \inf _{y} M(y)$. Integrating by parts, we have

$$\begin{aligned} \begin{aligned}&\int _K^{x- 1} \frac{{\overline{\nu }}_\lambda (x-y) }{{\overline{\nu }}_\lambda (x)} \mathrm {d}\nu _\lambda (y) \le C \int _K^{x-1} \left( \frac{x}{x-y} \right) ^{\alpha } \mathrm {d}\nu _\lambda (y) \\&= C x^\alpha \left[ \frac{{\overline{\nu }}_\lambda (K) }{(x-K)^\alpha } - {\overline{\nu }}_\lambda (x-1) + \int _K^{x-1} {\overline{\nu }}_\lambda (y) \alpha (x-y)^{-\alpha -1} \mathrm {d}y \right] . \end{aligned} \end{aligned}$$

(55)

The first term is small. The uniform continuity of M on compact sets, and its strict positivity implies that there is $\delta '>0$ small enough, such that for all $y \in [c^{-1/\alpha }, c^{1/\alpha }]$

$$\begin{aligned} \begin{aligned} (1- \delta ) M(y) \le \inf _{0 \le u \le \delta '} M( (1-u) y ) \le \sup _{0 \le u \le \delta '} M( (1-u) y ) \le (1 + \delta ) M(y). \end{aligned} \end{aligned}$$

Thus, using also that $\int _{\delta '}^1 u^{-\alpha -1} (1-u)^{-\alpha } \mathrm {d}u < \infty$, we obtain

$$\begin{aligned} \begin{aligned}&x^{\alpha } \int _K^{x-1} {\overline{\nu }}_\lambda (y) \alpha (x-y)^{-\alpha -1} \mathrm {d}y \\&= \frac{\alpha }{x^{\alpha } M(\lambda ^{1/\alpha })} \int _{1/x}^{1- K/x} \frac{M\left( (1-u) c^{\alpha ^{-1} \{\log _c x^\alpha \} }\lambda ^{1/\alpha } \right) }{u^{1+\alpha } (1-u)^{\alpha }} \mathrm {d}u \\&= \frac{\alpha }{x^{\alpha } M(\lambda ^{1/\alpha })} \int _{x^{-1}}^{\delta '} \frac{M \left( (1-u) c^{\alpha ^{-1} \{\log _c x^\alpha \} } \lambda ^{1/\alpha } \right) }{u^{\alpha +1} (1-u)^{\alpha }} \mathrm {d}u + O(x^{-\alpha }) \\&\le \frac{1 + \delta }{(1-\delta ')^\alpha } \frac{M(x \lambda ^{1/\alpha })}{M(\lambda ^{1/\alpha })} + O(x^{-\alpha }). \end{aligned} \end{aligned}$$

The lower bound follows similarly. Substituting back into (55)

$$\begin{aligned} \limsup _{x \rightarrow \infty } \sup _{\lambda \in [c^{-1},1]} \int _K^{x- 1} \frac{{\overline{\nu }}_\lambda (x-y) }{{\overline{\nu }}_\lambda (x)} \mathrm {d}\nu _\lambda (y) \le C \left[ \delta + \max \left\{ \frac{1+\delta }{(1- \delta ')^\alpha } -1 , \delta \right\} \right] . \end{aligned}$$

Since $\delta > 0$ and $\delta '>0$ are arbitrarily small, the statement follows.

1.3 A Technical Result Used in the Proof of Theorem 3

Lemma 6

Put $g = 1_{[\mathrm{e}^{-1},1]}$. For any $\delta > 0$ and $\varepsilon > 0$ there exist polynomials $Q_1$ and $Q_2$ such that

$$\begin{aligned} Q_1(x) \le g(x) \le Q_2(x), \quad x \in [0,1], \end{aligned}$$

and for any measure $\mu$ on $(0,\infty )$ such that $\int _0^\infty \mathrm{e}^{-x} \mu (\mathrm {d}x) < \infty$,

$$\begin{aligned} \begin{aligned}&\int _0^\infty \left[ Q_2(\mathrm{e}^{-x}) - g(\mathrm{e}^{-x} ) \right] \mu (\mathrm {d}x)< \varepsilon \int _0^\infty \mathrm{e}^{-x} \mu (\mathrm {d}x) + \mu ((1-\delta , 1+\delta )) \\&\int _0^\infty \left[ g(\mathrm{e}^{-x}) - Q_1(\mathrm{e}^{-x}) \right] \mu (\mathrm {d}x) < \varepsilon \int _0^\infty \mathrm{e}^{-x} \mu (\mathrm {d}x) + \mu ((1-\delta , 1+\delta )). \end{aligned} \end{aligned}$$

Proof

Fix $\varepsilon > 0$ and $\delta > 0$. Let

$$\begin{aligned} g_2 (x) = {\left\{ \begin{array}{ll} 0, &{} x \le \mathrm{e}^{-1} - \delta ', \\ \frac{e}{\delta '} (x - \mathrm{e}^{-1} + \delta '), &{} x \in [\mathrm{e}^{-1} - \delta ', \mathrm{e}^{-1}], \\ x^{-1}, &{} x \in [\mathrm{e}^{-1}, 1], \end{array}\right. } \end{aligned}$$

where $\delta ' > 0$ is chosen such that $- \log (\mathrm{e}^{-1} - \delta ') < 1 + \delta$. Then, $g_2$ is a continuous function on [0, 1], and $x g_2(x) \ge g(x)$. By the approximation theorem of Weierstrass, there is a polynomial $r_2(x)$, such that

$$\begin{aligned} \sup _{x \in [0,1]} | r_2(x) - (g_2(x) + \varepsilon /2) | \le \frac{\varepsilon }{2}. \end{aligned}$$

Let $Q_2(x) = x r_2(x)$. By the choice of $r_2$, we have $0 \le Q_2(x) - x g_2(x) \le \varepsilon x$. Moreover,

$$\begin{aligned} \begin{aligned} Q_2(x) - g(x) = Q_2(x) - x g_2(x) + x g_2(x) - g(x) \le \varepsilon x + 1_{[\mathrm{e}^{-1} - \delta ', \mathrm{e}^{-1}]}(x). \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} \int _0^\infty \left[ Q_2(\mathrm{e}^{-x}) - g(\mathrm{e}^{-x}) \right] \mu (\mathrm {d}x)&\le \varepsilon \int _0^\infty \mathrm{e}^{-x} \mu (\mathrm {d}x) + \mu \left( [1, - \log (\mathrm{e}^{-1} - \delta ')] \right) \\&\le \varepsilon \int _0^\infty \mathrm{e}^{-x} \mu (\mathrm {d}x) + \mu ( [1, 1+ \delta )). \end{aligned} \end{aligned}$$

The construction of $Q_1$ is similar. Choose

$$\begin{aligned} g_1(x) = {\left\{ \begin{array}{ll} 0, &{} x \le \mathrm{e}^{-1}, \\ (\delta ')^{-1} (\mathrm{e}^{-1} + \delta ')^{-1} (x-\mathrm{e}^{-1}), &{} x \in [\mathrm{e}^{-1}, \mathrm{e}^{-1}+\delta '], \\ x^{-1}, &{} x \ge \mathrm{e}^{-1} + \delta ', \end{array}\right. } \end{aligned}$$

and let $r_1$ be a polynomial such that

$$\begin{aligned} \sup _{x \in [0,1]} | r_1(x) - (g_1(x) -\varepsilon /2)| \le \frac{\varepsilon }{2}. \end{aligned}$$

The same proof shows that $Q_1(x) = x r_1(x)$ satisfies the stated properties. $\square$

1.4 Verifying that (40) Holds

To ease notation put $a = {2\pi \alpha }/{\log c}$. Recall (35). For the first derivative

$$\begin{aligned} \xi '(n) = -\frac{1}{2n^{\alpha +1}} \left( \alpha (1+ 2\varepsilon \sin (a \log n))- 2 \varepsilon a \cos (a \log n) \right) < 0, \end{aligned}$$

whenever $\varepsilon$ is small enough. Long but straightforward calculation gives

$$\begin{aligned} \varDelta \xi _n = \xi _{n} - \xi _{n-1} = \frac{n^{-\alpha -1}}{2} \left[ x_0(n) + \frac{x_1(n)}{n} + \frac{x_2(n)}{n^2} + O(n^{-3}) \right] , \end{aligned}$$

(56)

where $x_0(n) = \alpha ( 1 + 2 \varepsilon \sin ( a \log n) ) - 2 \varepsilon a \cos (a \log n)$, and the first- and second-order terms are $x_i(n) = c^i_0 + c^i_1 \sin (a \log n) + c^i_2 \cos (a \log n)$, $i = 1,2,$ where $c^i_j$ are constants, whose actual value is not important for us. Note that $x_0(n)$ comes from the first derivative, and we use it frequently that

$$\begin{aligned} x_0(n) \ge \alpha - 2 \varepsilon (a + \alpha ) > 0 \end{aligned}$$

(57)

for $\varepsilon > 0$ small enough. From (56), we deduce that

$$\begin{aligned} \frac{\varDelta \xi _{n-2}}{\varDelta \xi _{n-1}} = 1 + \frac{\alpha _n}{n} + \frac{r_n}{n^2} + R_n, \end{aligned}$$

with $R_n = O(n^{-3})$, and $\alpha _n = H_1( a \log n)$, $r_n = H_2( a \log n)$, where

$$\begin{aligned} \begin{aligned} H_1(x)&= 1 + \alpha - 2 \varepsilon a \frac{a \sin x + \alpha \cos x}{\alpha ( 1 + 2 \varepsilon \sin x) - 2 \varepsilon a \cos x} \\ H_2(x)&= \frac{a_0^{2} + a_1^{2} \sin x + a_2^2 \sin (2x) + b_1^{2} \cos x + b_2^{2} \cos (2x)}{(\alpha ( 1 + 2 \varepsilon \sin x) - 2 \varepsilon a \cos x)^2} \end{aligned} \end{aligned}$$

(58)

with some constants $a_j^{2}, b_j^{2}$, whose value is not important. By (57), the denominators in $H_1, H_2$ are strictly positive, and therefore, $H_1$ and $H_2$ are continuous smooth ($C^{\infty }$) functions. This implies that $\alpha _n = 1 + \alpha + O(\varepsilon )$, $r_n=O(1)$,

$$\begin{aligned} \alpha _n - \alpha _{n-1} = O(n^{-1}), \ \alpha _{n-1} + \alpha _{n+1} - 2 \alpha _n= O(n^{-2}), \ r_n - r_{n-1} = O(n^{-1}). \end{aligned}$$

This is everything we need for the construction of $f_\varepsilon$ in Sect. 6.1.

1.5 Distortion Properties for F

Let $J_n := [ (\xi _n+1)/2, (\xi _{n-1}+1)/2 )$ be the intervals on which $F := F_\varepsilon$ is continuous.

Lemma 7

There exists $K>0$ such that $\frac{F''(x)}{F'(x)^2} \le K$ for all n and all $x \in J_n$. In particular, $F|_{J_n}$ can be extended to $\overline{J_n}$ for each n so that $\frac{F''(x)}{F'(x)^2} \le K$ for all $x \in {\overline{J}}_n$.

Proof

Given the map $f_{\varepsilon , n}$ in Sect. 6.1, it is easy to see that for $x \in [\xi _{n-1}, \xi _n]$, $n\ge 1$,

$$\begin{aligned} f_{\varepsilon ,n}''(x) = \frac{A_n}{\xi _{n-1}-\xi _n} = O(n^{\alpha -1}) \quad \text { and } \quad \left| f_{\varepsilon , n}'(x)- \left( 1+\frac{\alpha _n}{n} \right) \right| = O(n^{-2}), \end{aligned}$$

where the derivatives at the end-points are interpreted as one-sided derivatives. From (58) at the end of the previous subsection, we know that $\alpha _n = 1+\alpha + O(\varepsilon )$ as $n \rightarrow \infty$ and $\varepsilon \rightarrow 0$. Since $\alpha > 0$, we can choose $\delta > 0$ small enough such that $(1 + \alpha ) (1 - \delta ) > 1$. For n large enough $f_{\varepsilon ,n}'(x) \ge 1+ \frac{1}{n}(1+\alpha )(1-\delta ) > 1$. It follows that $D(f_{\varepsilon ,n}) := {f_{\varepsilon ,n}''}/{(f_{\varepsilon ,n}')^2}$ is uniformly bounded. Next, compute that for any two $C^2$ functions g, h,

$$\begin{aligned} D(g \circ h) = D(g) \circ h + \frac{1}{g'\circ h} D(h). \end{aligned}$$

Applying this to $g = f^{n-1}$ and $h = f$, gives

$$\begin{aligned} D(f^n) = D(f^{n-1}) \circ f + \frac{1}{(f^{n-1})' \circ f} D(f). \end{aligned}$$

Write $x_k = f_\varepsilon ^k(x)$ for $k \ge 0$. For some $C = C(\delta ) > 0$

$$\begin{aligned} (f_\varepsilon ^{n-1})'(x)= & {} f_\varepsilon '(x_{n-2}) f_\varepsilon '(x_{n-3}) \cdots f_\varepsilon '(x_0) \\\ge & {} C \left( 1+\frac{(1+\alpha )(1- \delta )}{n-1} \right) \left( 1+\frac{(1+\alpha )(1- \delta )}{n-2} \right) \cdots 2 \\= & {} 2 C \exp \left( \sum _{k=2}^n \log \left[ 1+\frac{(1+\alpha )(1- \delta )}{k-1} \right] \right) \\\sim & {} 2 C \exp (((1+\alpha )(1- \delta ))\log n + C_n) \\\ge & {} C' n^{(1+\alpha )(1- \delta )}, \end{aligned}$$

where $(C_n)$ is a bounded sequence and $C' >0$. We get

$$\begin{aligned} D(f^n) \le D(f^{n-1}) \circ f + \frac{1}{C' n^{(1+\alpha )(1- \delta )} } D(f). \end{aligned}$$

By induction,

$$\begin{aligned} D(F|_{J_n}) \le D(f^n|_{J_n}) \ll D(f) \sum _{k=2}^{n-1} \frac{1}{C' k^{(1+\alpha )(1- \delta )}}, \end{aligned}$$

which is bounded in n since the exponent $(1+\alpha )(1- \delta ) > 1$. $\square$