Weighted laws of large numbers and convergence of weighted ergodic averages on vector valued $L_p$-spaces

Abstract

A more general notion of weight called admissible is introduced and then an investigation is carried out on the a.e. convergence of weighted strong laws of large numbers and their applications to weighted ergodic averages on vector-valued $L_p$-spaces.

Appendices

Appendix A: Examples of admissible weighted

In this section, we provide some examples of admissible weights (see [40]).

Example A.1

Let $G_n=n^{1-\beta }$, for some $\beta \in (0,1]$ and $p>1$. Define a weight $\{W_n\}$ by $W_n=n^{1-\delta }$ with $\delta \in [0,\frac{p-1}{p}\beta )$. Let us set $n_k=[k^r]+1$, where $r=1/\beta$. Then, $\{W_n\}$ is $(p,\{n_k\})$-admissible. Indeed, one has

$$\begin{aligned} \sum _{k=1}^\infty \left( \frac{G_{n_k}}{W_{n_k}}\right) ^p&= {} \sum _{k=1}^\infty \frac{1}{n_k^{(\beta -\delta )p}}\\ &\le {} \sum _{k=1}^\infty \frac{1}{k^{pr(\beta -\delta )}}<\infty \end{aligned}$$

since $pr(\beta -\delta )=(1-r\delta )p>1$. Hence, the condition (2.1) is satisfied. Let us check (2.2). From the definition of the sequences, one gets

$$\begin{aligned} \frac{n_{k+1}-n_k}{W_{n_k}}&\le {} \frac{(k+1)^r+1-k^r}{(k^r)^{1-\delta }} \\ &\le {} \frac{2r(k+2)^{r-1}}{k^{r(1-\delta )}} \\&= {} 2r\left( \frac{k+2}{k}\right) ^{r-1}\frac{1}{k^{1-r\delta }} \end{aligned}$$

(A.1)

Hence,

$$\begin{aligned} \left( \frac{n_{k+1}-n_k}{W_{n_k}}\right) ^p\le (2r)^p\left( \frac{k+2}{k}\right) ^{p(r-1)}\frac{1}{k^{p(1-r\delta )}} \end{aligned}$$

(A.2)

which, due to the choice of $\delta$, implies (2.2). We point out that this kind of weight was considered in [13]

Example A.2

Let ${{\tilde{G}}}_n=n^{1-\beta }\ln ^\gamma n$, for some $\beta \in (0,1]$, $\gamma \ge 1$ and $p>1$. Then a weight given by ${{\tilde{W}}}_n=n^{1-\delta }\ln ^\gamma n$ with $\delta \in [0,\frac{p-1}{p}\beta )$ is $(p,\{n_k\})$-admissible w.r.t. $\{{{\tilde{G}}}_n\}$, here as before, $n_k=[k^r]+1$, where $r=1/\beta$. Using the same argument as in Example A.1, we easily obtain (2.1). Now according to (A.2), one finds

$$\begin{aligned} \left( \frac{n_{k+1}-n_k}{{{\tilde{W}}}_{n_k}}\right) ^p &\le {} (2r)^p\left( \frac{k+2}{k}\right) ^{p(r-1)}\frac{1}{k^{p(1-r\delta )}\ln ^{\gamma p} k}\\ &\le {} (2r)^p\left( \frac{k+2}{k}\right) ^{p(r-1)}\frac{1}{k^{p(1-r\delta )}} \end{aligned}$$

and we arrive at (2.2).

Example A.3

Let $R_n=\frac{n^{1-\beta }}{\ln ^\alpha n}$, for some $\beta \in (0,1]$, $\alpha \ge 1$ and $p>1$. Then a weight given by $U_n=\frac{n^{1-\delta }}{\ln ^\alpha n}$ with $\delta \in [0,\frac{p-1}{p}\beta )$ is $(p,\{n_k\})$-admissible w.r.t. $\{R_n\}$, where $n_k=[k^r]+1$, $r=1/\beta$. One can check that (2.1) is easily satisfied. Again using (A.2), we obtain

$$\begin{aligned} \frac{n_{k+1}-n_k}{ U_{n_k}}&\le {} \frac{2r(k+2)^{r-1}\ln ^\alpha n_k}{k^{r(1-\delta )}}\\ &\le {} \frac{2r(k+2)^{r-1}(r+1)^\alpha \ln ^\alpha k}{k^{r(1-\delta )}}\\&= {} 2r(r+1)^\alpha \left( \frac{k+2}{k}\right) ^{r-1}\frac{\ln ^\alpha k}{k^{r(1-\delta )}}. \end{aligned}$$

One can check that the series

$$\begin{aligned} \sum _{k=1}^\infty \frac{\ln ^{\alpha p} k}{k^{pr(1-\delta )}} \end{aligned}$$

converges, if $pr(1-\delta )>1$. Hence, the condition (2.2) is satisfied.

Example A.4

Now we are going to provide a more general example. Let $\{G_n\}$ be a weight such that

$$\begin{aligned} \sum _{k=1}^\infty \frac{1}{(G_k)^{p\epsilon }}<\infty \end{aligned}$$

(5.3)

for some $\epsilon >0$ and $p>1$. Then a sequence $\{W_n\}$ is given by

$$\begin{aligned} W_n=(G_n)^{\epsilon +1}, \quad n\in {\mathbb {N}}. \end{aligned}$$

Now, define the sequence $\{n_k\}$ as follows:

$$\begin{aligned} n_1=1, \quad n_{k+1}=[G_{n_k}]+n_k+1, \quad k\in {\mathbb {N}}. \end{aligned}$$

Then $\{W_n\}$ is $(p,\{n_k\})$-admissible w.r.t. $\{G_n\}$. Indeed,

$$\begin{aligned} \sum _{k=1}^\infty \left( \frac{G_{n_k}}{W_{n_k}}\right) ^p=\sum _{k=1}^\infty \frac{1}{(G_{n_k})^{p\epsilon }}<\infty \end{aligned}$$

which yields (2.1). Let us check the second condition. One has

$$\begin{aligned} \frac{n_{k+1}-n_k}{{{\tilde{W}}}_{n_k}}&= {} \frac{[G_{n_{k}}]+1}{(G_{n_k})^{1+\epsilon }}\\ &\sim {} \frac{1}{(G_{n_k})^{\epsilon }}+ \frac{1}{(G_{n_k})^{\epsilon +1}}\\ &\le {} \frac{1}{(G_{n_k})^{\epsilon }}. \end{aligned}$$

This implies the condition (2.2).

For instance, if we takes $G_n=n^{1/p}(\ln n)^{\beta +1/p}(\ln \ln n)^\gamma$ (with $\beta ,\gamma >0$, $p>1$), then

$$\begin{aligned} \sum _{k=1}^\infty \frac{1}{(G_n)^{p}}<\infty \end{aligned}$$

hence, $W_n=G_n^2$ is $(p,\{n_k\})$-admissible.

Appendix B: More examples with condition (3.8)

Let us consider some examples for which all the above noticed conditions are satisfied.

Example B.1

Let $\{G_n\}$ and $\{W_n\}$ be weights considered in Example A.1, i.e. $G_n=n^{1-\beta }$, $W_n=n^{1-\delta }$, where $\beta \in (0,1]$, $p>1$ and $\delta \in [0,\frac{p-1}{p}\beta )$. One can see that

$$\begin{aligned} \frac{G_k}{W_k}=\frac{1}{k^{\beta -\delta }} \end{aligned}$$

due to $\beta -\delta <1$, we infer $\sum _{k=1}^\infty \frac{G_k}{W_k}=\infty$, while (3.8) is satisfied. Indeed,

$$\begin{aligned} \frac{G_k}{W_k}\left( 1-\frac{W_k}{W_{k+1}}\right)&= {} \frac{1}{k^{\beta -\delta }}\frac{(k+1)^{1-\delta }-k^{1-\delta }}{(k+1)^{1-\delta }}\\ &\le {} \frac{1}{k^{\beta -\delta }}\frac{(1-\delta )k^{-\delta }}{k^{1-\delta }}\\&= {} \frac{1-\delta }{k^{1+\beta -\delta }} \end{aligned}$$

One can see that $1+\beta -\delta >1$, therefore,

$$\begin{aligned} \sum _{k=1}^\infty \frac{G_k}{W_k}\left( 1-\frac{W_k}{W_{k+1}}\right) <\infty . \end{aligned}$$

Example B.2

Let $\{R_n\}$ and $\{U_n\}$ be weights considered in Example A.3, i.e. $R_n=\frac{n^{1-\beta }}{\ln ^\alpha n}$, $U_n=\frac{n^{1-\delta }}{\ln ^\alpha n}$, where $\beta \in (0,1]$, $\alpha \ge 1$, $p>1$ and $\delta \in [0,\frac{p-1}{p}\beta )$. It is clear that

$$\begin{aligned} \sum _{k=1}^\infty \frac{R_k}{U_k}=\infty . \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \frac{R_k}{U_k}\left( 1-\frac{U_k}{U_{k+1}}\right)&= {} \frac{1}{k^{\beta -\delta }}\left( 1-\frac{k^{1-\delta }}{(k+1)^{1-\delta }} \frac{\ln ^\alpha (k+1)}{\ln ^\alpha k}\right) \\ &\le {} \frac{1}{k^{\beta -\delta }}\left( 1-\frac{k^{1-\delta }}{(k+1)^{1-\delta }}\right) \\ &\le {} \frac{1-\delta }{k^{1+\beta -\delta }} \end{aligned}$$

Hence, the series

$$\begin{aligned} \sum _{k=1}^\infty \frac{R_k}{U_k}\left( 1-\frac{U_k}{U_{k+1}}\right) \end{aligned}$$

converges.

Example B.3

Let us provide an other example of p-admissible weight. We are going to establish that the weight considered in Example 2.1 is indeed p-admissible.

Let $G_n=(\ln n)^{\beta +1/p}(\ln \ln n)^\gamma$ with $\beta >0$, $\gamma \ge 1$, and $n_m=[m^r]+1$, $r=1/\alpha$ with $\alpha \in (0,1]$. Define

$$\begin{aligned} W_n=n^{\delta /p}(\ln n)^{\beta +1/p}(\ln \ln n)^\gamma , \quad \delta \ge p(1-\alpha )+1. \end{aligned}$$

Then $\{W_n\}$ is p-admissible w.r.t. $\{G_n\}$. Indeed, we first note that

$$\begin{aligned} \sum _{m=1}^\infty \frac{1}{n_m}<\infty \end{aligned}$$

therefore

$$\begin{aligned} \sum _{k=1}^\infty \left( \frac{G_{n_m}}{W_{n_m}}\right) ^p=\sum _{m=1}^\infty \frac{1}{n_m^{\delta }}<\infty . \end{aligned}$$

Furthermore, using the same argument of Example A.1

$$\begin{aligned} \frac{n_{m+1}-n_m}{W_{n_m}}&= \frac{n_{m+1}-n_m}{n_m^{\delta /p}G_{n_m}}\\ &\le {} \frac{(m+1)^r+1-m^r}{m^{(1-\tau )r}G_{n_m}} \\&= {} 2r\left( \frac{m+2}{m}\right) ^{r-1}\frac{1}{m^{1-r\tau }G_{n_m}}, \end{aligned}$$

where $\tau =(p-\delta )/p$. Hence,

$$\begin{aligned} \left( \frac{n_{m+1}-n_m}{W_{n_m}}\right) ^p &\sim {} \frac{1}{m^{(1-r\tau )p}G^p_{n_m}}\\ &\sim {} \frac{1}{m^{(1-r\tau )p}(\ln m)^{\beta p+1}(\ln \ln m)^{\gamma p}}. \end{aligned}$$

Consequently, if $(1-r\tau )p\ge 1$, i.e. $\delta \ge p(1-\alpha )+1$, the series

$$\begin{aligned} \sum _{m=1}^\infty \frac{1}{m^{(1-r\tau )p}(\ln n_m)^{\beta p+1}(\ln \ln n_m)^{\gamma p}} \end{aligned}$$

converges. So, $\{W_n\}$ is p-admissible w.r.t. $\{G_n\}$.

Let us establish that the weights $\{G_n\}$ and $\{W_n\}$ satisfy the condition (3.8). For the sake of simplicity, we assume that $\delta =1$. Then

$$\begin{aligned} \frac{G_k}{W_k}\left( 1-\frac{W_k}{W_{k+1}}\right)&= {} \frac{(k+1)^{1/p}G_{k+1}-k^{1/p}G_{k}}{k^{1/p}(k+1)^{1/p}G_{k+1}} \end{aligned}$$

(B.1)

Now considering the derivative of $\psi (x)=x^{1/p}(\ln x)^{\beta +1/p}(\ln \ln x)^\gamma$, we can estimate RHS of (B.1) as follows

$$\begin{aligned} \frac{(k+1)^{1/p}G_{k+1}-k^{1/p}G_{k}}{k^{1/p}(k+1)^{1/p}G_{k+1}}\le \frac{C}{k^{1+1/p}}, \quad k\in {\mathbb {N}} \end{aligned}$$

for some constant. This implies the convergence of the series

$$\begin{aligned} \sum _{k=1}^\infty \frac{G_k}{W_k}\left( 1-\frac{W_k}{W_{k+1}}\right) . \end{aligned}$$