A note on the Analogue of Lebesgue-Radon-Nikodym theorem with respect to weighted p-adic q-measure on ${\mathbb{Z}}_p$

Abstract

We give the analogue of the Lebesgue-Radon-Nikodym theorem with respect to a weighted p-adic q-measure on ${\mathbb{Z}}_p$. In a special case, when the weight $q^x$ is 1, we can derive the same result as Kim et al. (Abstr. Appl. Anal. 2011:637634, 2011). And if $q=1$, we have the same result as Kim (Russ. J. Math. Phys. 19:193-196, 2012).

MSC:11B68, 11S80.

1 Introduction

Let p be a fixed odd prime number. Throughout this paper, the symbols ${\mathbb{Z}}_p$, ${\mathbb{Q}}_p$, and ${\mathbb{C}}_p$ denote the ring of p-adic integers, the field of p-adic rational numbers, and the p-adic completion of the algebraic closure of ${\mathbb{Q}}_p$, respectively. Let ${\nu }_p$ be the normalized exponential valuation of ${\mathbb{C}}_p$ with $|p|=p^{-{\nu }_p(p)}=\frac{1}{p}$ and ${\nu }_p(0)=\mathrm{\infty }$.

When one speaks of q-extension, q can be regarded as an indeterminate, a complex $q\in \mathbb{C}$, or a p-adic number $q\in {\mathbb{C}}_p$. In this paper, we assume that $q\in {\mathbb{C}}_p$ with $|1-q|<1$, and we use the notations of q-numbers as follows:

$${[x]}_q=[x:q]=\frac{1-q^x}{1-q}\text{and}{[x]}_{-q}=\frac{1-{(-q)}^x}{1+q}.$$

(1.1)

For any positive integer N, let

$$a+p^N{\mathbb{Z}}_p=\{x\in {\mathbb{Z}}_p|x\equiv a(mod{p}^N)\},$$

(1.2)

where $a\in \mathbb{Z}$ satisfies the condition $0\le a<p^N$ (see [1–8]).

It is known that the fermionic p-adic q-measure on ${\mathbb{Z}}_p$ is given by Kim as follows:

$${\mu }_{-q}(a+p^N{\mathbb{Z}}_p)=\frac{{(-q)}^a}{{[p^N]}_{-q}}=\frac{1+q}{1+q^{p^N}}{(-q)}^a,\text{(see [1, 6, 9–12])}.$$

(1.3)

Let $C({\mathbb{Z}}_p)$ be the space of continuous functions on ${\mathbb{Z}}_p$. From (1.3), the fermionic p-adic q-integral on ${\mathbb{Z}}_p$ is defined by Kim as follows:

$$I_{-q}(f)={\int }_{{\mathbb{Z}}_p}f(x)d{\mu }_{-q}(x)=\underset{N\to \mathrm{\infty }}{lim}\frac{1}{{[p^N]}_{-q}}\sum _{x=0}^{p^N-1}f(x){(-q)}^x,$$

(1.4)

where $f\in C({\mathbb{Z}}_p)$ (see [1, 6, 9–12]).

Let us assume $q\in {\mathbb{C}}_p$ with $|q-1|<1$. By (1.4), we get

$${\int }_{{\mathbb{Z}}_p}{q}^{-x}{e}^{{[x]}_{q}t}d{\mu }_{-q}(x)=\sum _{n=0}^{\mathrm{\infty }}{E}_{n,q}\frac{t^n}{n!},$$

(1.5)

(see [7, 8, 13]) where $E_{n,q}$ are q-Euler numbers. The q-Euler polynomials, $E_{n,q}(x)$, are also defined by

$${\int }_{{\mathbb{Z}}_p}{q}^{-y}{e}^{{[x+y]}_{q}t}d{\mu }_{-y}(t)=\sum _{n=0}^{\mathrm{\infty }}{E}_{n,q}(x)\frac{t^n}{n!}.$$

(1.6)

By (1.5) and (1.6), we get

$$E_{n,q}(x)=\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)x^{n-l}{E}_{l,q}={(x+E_q)}^n,$$

with the usual convention of replacing ${(E_q)}^n$ by $E_{n,q}$ (see [1, 2, 7, 8, 13]),

$$E_{n,q}={\int }_{{\mathbb{Z}}_p}{q}^{-x}{[x]}_q^{n}d{\mu }_{-q}(x)=\frac{{[\iota ]}_q}{{(1-q)}^n}\sum _{l=0}^n{(-1)}^l\left(\begin{array}{c}n\\ l\end{array}\right)\frac{1}{{[l]}_q}.$$

We will give the analogue of the Lebesgue-Radon-Nikodym theorem with respect to a weighted p-adic q-measure on ${\mathbb{Z}}_p$. In a special case, when the weight $q^x$ is 1, we can derive the same result as Kim et al. [9]. And if $q=1$, we have the same result as Kim [4].

2 Lebesgue-Radon-Nikody-type theorem with respect to a weighted p-adic q-measure on ${\mathbb{Z}}_p$

For any positive integer a and n, with $a<p^n$ and $f\in C({\mathbb{Z}}_p)$, let us define

$${\tilde{\mu }}_{f,-q}(a+p^n{\mathbb{Z}}_p)={\int }_{a+p^n{\mathbb{Z}}_p}{q}^{-x}f(x)d{\mu }_{-q}(x),$$

(2.1)

where the integral is the fermionic p-adic q-integral on ${\mathbb{Z}}_p$.

From (1.3), (1.4), and (2.1), we note that

(2.2)

By (2.2), we get

$${\tilde{\mu }}_{f,-q}(a+p^n{\mathbb{Z}}_p)=\frac{{[2]}_q}{{[2]}_{q^{p^n}}}{(-1)}^a{\int }_{{\mathbb{Z}}_p}{q}^{p^{n}x}f(a+p^{n}x)d{\mu }_{-q^{p^n}}(x).$$

(2.3)

Therefore, by (2.3), we obtain the following theorem.

Theorem 1 For $f,g\in C({\mathbb{Z}}_p)$, we have

$${\tilde{\mu }}_{\alpha f+\beta g,-q}=\alpha {\tilde{\mu }}_{f,-q}+\beta {\tilde{\mu }}_{g,-q},$$

(2.4)

where α, β are constants.

From (2.2) and (2.4), we note that

$$\left|{\tilde{\mu }}_{f,-q}(a+p^n{\mathbb{Z}}_p)\right|\le M{\parallel f_q\parallel }_{\mathrm{\infty }},$$

(2.5)

where ${\parallel f_q\parallel }_{\mathrm{\infty }}={sup}_{x\in {\mathbb{Z}}_p}|q^{-x}f(x)|$ and M is some positive constant.

Now, we recall the definition of the strongly fermionic p-adic q-measure on ${\mathbb{Z}}_p$. If ${\mu }_{-q}$ satisfies the following equation:

$$|{\mu }_{-q}(a+p^n{\mathbb{Z}}_p)-{\mu }_{-q}(a+p^{n+1}{\mathbb{Z}}_p)|\le {\delta }_{n,q},$$

(2.6)

where ${\delta }_{n,q}\to 0$ and $n\to \mathrm{\infty }$ and ${\delta }_{n,q}$ is independent of a, then ${\mu }_{-q}$ is called a weakly fermionic p-adic q-measure on ${\mathbb{Z}}_p$.

If ${\delta }_{n,q}$ is replaced by $C{p}^{-{\nu }_p(1-q^n)}$ (C is some constant), then ${\mu }_{-q}$ is called a strongly fermionic p-adic q-measure on ${\mathbb{Z}}_p$.

Let $P(x)\in {\mathbb{C}}_p[{[x]}_q]$ be an arbitrary q-polynomial with $\sum a_i{[x]}_q^i$. Then we see that ${\mu }_{P,-q}$ is a strongly fermionic p-adic q-measure on ${\mathbb{Z}}_p$. Without loss of generality, it is enough to prove the statement for $P(x)={[x]}_q^k$.

Let a be an integer with $0\le a<p^n$. Then we get

$${\tilde{\mu }}_{P,-q}(a+p^n{\mathbb{Z}}_p)=\frac{{[2]}_q}{{[2]}_{q^{p^n}}}{(-1)}^a\underset{m\to \mathrm{\infty }}{lim}\frac{1}{{[p^{m-n}]}_{-q^{p^n}}}\sum _{i=0}^{p^{m-n}-1}{[a+i{p}^n]}_q^k{(-1)}^i{q}^{p^{n}i},$$

(2.7)

and

$$q^{p^{n}i}=\sum _{l=0}^i\left(\genfrac{}{}{0ex}{}{i}{l}\right){\left[p^n\right]}_q^l{(q-1)}^l.$$

By (2.7), we easily get

$$\begin{array}{rl}{\tilde{\mu }}_{P,-q}(a+p^n{\mathbb{Z}}_p)& \equiv \frac{{[2]}_q}{{[2]}_{q^{p^n}}}{(-1)}^a{[a]}_q^k(mod{\left[p^n\right]}_q)\\ \equiv \frac{{[2]}_q}{{[2]}_{q^{p^n}}}{(-1)}^{a}P(a)(mod{\left[p^n\right]}_q).\end{array}$$

(2.8)

Let x be arbitrary in ${\mathbb{Z}}_p$ with $x\equiv x_n(mod{p}^n)$ and $x\equiv x_{n+1}(mod{p}^{n+1})$, where $x_n$ and $x_{n+1}$ are positive integers such that $0\le x_n<p^n$ and $0\le x_{n+1}<p^{n+1}$. Thus, by (2.8), we have

$$|{\tilde{\mu }}_{P,-q}(a+p^n{\mathbb{Z}}_p)-{\tilde{\mu }}_{P,-q}(a+p^{n+1}{\mathbb{Z}}_p)|\le C{p}^{-{\nu }_p(1-q^{p^n})},$$

(2.9)

where C is some positive constant and $n\gg 0$.

Let

$$f_{{\tilde{\mu }}_{P,-q}}(a)=\underset{n\to \mathrm{\infty }}{lim}{\tilde{\mu }}_{P,-q}(a+p^n{\mathbb{Z}}_p).$$

(2.10)

Then by (2.5), (2.7), and (2.8), we get

$$f_{{\tilde{\mu }}_{P,-q}}(a)=\frac{{[2]}_q}{2}{(-1)}^a{[a]}_q^k=\frac{{[2]}_q}{2}{(-1)}^{a}P(a).$$

(2.11)

Since $f_{{\tilde{\mu }}_{P,-q}}(x)$ is continuous on ${\mathbb{Z}}_p$, it follows, for all $x\in {\mathbb{Z}}_p$,

$$f_{{\tilde{\mu }}_{P,-q}}(x)=\frac{{[2]}_q}{2}{(-1)}^{x}P(x).$$

(2.12)

Let $g\in C({\mathbb{Z}}_p)$. By (2.10), (2.11), and (2.12), we get

$$\begin{array}{rcl}{\int }_{{\mathbb{Z}}_p}g(x)d{\tilde{\mu }}_{P,-q}(x)& =& \underset{n\to \mathrm{\infty }}{lim}\sum _{i=0}^{p^n-1}g(i){\tilde{\mu }}_{P,-q}(i+p^n{\mathbb{Z}}_p)\\ =& \frac{{[2]}_q}{2}\underset{n\to \mathrm{\infty }}{lim}\sum _{i=0}^{p^n-1}g(i){(-q)}^i{[i]}_q^k\\ =& {\int }_{{\mathbb{Z}}_p}{q}^{-x}g(x){[x]}_q^{k}d{\mu }_{-q}(x).\end{array}$$

(2.13)

Therefore, by (2.13), we obtain the following theorem.

Theorem 2 Let $P(x)\in {\mathbb{C}}_p[{[x]}_q]$ be an arbitrary q-polynomial with $\sum a_i{[x]}_q^i$. Then ${\tilde{\mu }}_{P,-q}$ is a strongly fermionic weighted p-adic q-measure on ${\mathbb{Z}}_p$, and for all $x\in {\mathbb{Z}}_p$,

$$f_{{\tilde{\mu }}_{P,-q}}={(-1)}^x\frac{{[2]}_q}{2}P(x).$$

(2.14)

Furthermore, for any $g\in C({\mathbb{Z}}_p)$, we have

$${\int }_{{\mathbb{Z}}_p}g(x)d{\tilde{\mu }}_{P,-q}(x)={\int }_{{\mathbb{Z}}_p}{q}^{-x}g(x)P(x)d{\mu }_{-q}(x),$$

(2.15)

where the second integral is a fermionic p-adic q-integral on ${\mathbb{Z}}_p$.

Let $f(x)={\sum }_{n=0}^{\mathrm{\infty }}{a}_{n,q}{\left(\genfrac{}{}{0ex}{}{x}{n}\right)}_q$ be the q-Mahler expansion of a continuous function on ${\mathbb{Z}}_p$, where

$${\left(\genfrac{}{}{0ex}{}{x}{n}\right)}_q=\frac{{[x]}_q{[x-1]}_q\cdots {[x-n+1]}_q}{{[n]}_q!}\left(\text{see [4]}\right).$$

(2.16)

Then we note that ${lim}_{n\to \mathrm{\infty }}|a_{n,q}|=0$.

Let

$$f_m(x)=\sum _{i=0}^m{a}_{i,q}{\left(\genfrac{}{}{0ex}{}{x}{i}\right)}_q\in {\mathbb{C}}_p\left[{[x]}_q\right].$$

(2.17)

Then

$${\parallel f-f_m\parallel }_{\mathrm{\infty }}\le \underset{m\le n}{sup}|a_{n,q}|.$$

(2.18)

Writing $f=f_m+f-f_m$, we easily get

(2.19)

From Theorem 2, we note that

$$\left|{\tilde{\mu }}_{f-f_m,-q}(a+p^n{\mathbb{Z}}_p)\right|\le {\parallel f-f_m\parallel }_{\mathrm{\infty }}\le C_1{p}^{-{\nu }_p(1-q^{p^n})},$$

(2.20)

where $C_1$ is some positive constant.

For $m\gg 0$, we have ${\parallel f\parallel }_{\mathrm{\infty }}={\parallel f_m\parallel }_{\mathrm{\infty }}$.

So,

$$|{\tilde{\mu }}_{f_m,-q}(a+p^n{\mathbb{Z}}_p)-{\tilde{\mu }}_{f_m,-q}(a+p^{n+1}{\mathbb{Z}}_p)|\le C_2{p}^{-{\nu }_p(1-q^{p^n})},$$

(2.21)

where $C_2$ is also some positive constant.

By (2.20) and (2.21), we see that

(2.22)

If we fix $ϵ>0$ and fix m such that $\parallel f-f_m\parallel \le ϵ$, then for $n\gg 0$, we have

$$|f(a)-{\tilde{\mu }}_{f,-q}(a+p^n{\mathbb{Z}}_p)|\le ϵ.$$

(2.23)

Hence, we have

$$f_{{\mu }_{f,-q}}(a)=\underset{n\to \mathrm{\infty }}{lim}{\tilde{\mu }}_{f,-q}(a+p^n{\mathbb{Z}}_p)=\frac{{[2]}_q}{2}{(-1)}^{a}f(a).$$

(2.24)

Let m be a sufficiently large number such that ${\parallel f-f_m\parallel }_{\mathrm{\infty }}\le p^{-n}$.

Then we get

$$\begin{array}{rcl}{\tilde{\mu }}_{f,-q}(a+p^n{\mathbb{Z}}_p)& =& {\tilde{\mu }}_{f_m,-q}(a+p^n{\mathbb{Z}}_p)+{\tilde{\mu }}_{f-f_m,-q}(a+p^n{\mathbb{Z}}_p)\\ =& {\tilde{\mu }}_{f_m,-q}(a+p^n{\mathbb{Z}}_p)\\ =& {(-1)}^a\frac{{[2]}_q}{{[2]}_{q^{p^n}}}f(a)(mod{\left[p^n\right]}_q).\end{array}$$

(2.25)

For any $g\in C({\mathbb{Z}}_p)$, we have

$${\int }_{{\mathbb{Z}}_p}g(x)d{\tilde{\mu }}_{f,-q}(x)={\int }_{{\mathbb{Z}}_p}{q}^{-x}f(x)g(x)d{\mu }_{-q}(x).$$

(2.26)

Assume that f is the function from $C({\mathbb{Z}}_p,{\mathbb{C}}_p)$ to $Lip({\mathbb{Z}}_p,{\mathbb{C}}_p)$. By the definition of ${\tilde{\mu }}_{-q}$, we easily see that ${\tilde{\mu }}_{-q}$ is a strongly p-adic q-measure on ${\mathbb{Z}}_p$, and for $n\gg 0$,

$$|f_{{\tilde{\mu }}_{-q}}(a)-{\tilde{\mu }}_{-q}(a+p^n{\mathbb{Z}}_p)|\le C_3{p}^{-{\nu }_p(1-q^{p^n})},$$

(2.27)

where $C_3$ is some positive constant.

If ${\tilde{\mu }}_{1,-q}$ is associated with strongly fermionic weighted p-adic q-measure on ${\mathbb{Z}}_p$, then we have

$$|{\tilde{\mu }}_{1,-q}(a+p^n{\mathbb{Z}}_p)-f_{{\tilde{\mu }}_{-q}}(a)|\le C_4{p}^{-{\nu }_p(1-q^{p^n})},$$

(2.28)

where $n\gg 0$ and $C_4$ is some positive constant.

From (2.28), we get

(2.29)

where K is some positive constant.

Therefore, ${\tilde{\mu }}_{-q}-{\tilde{\mu }}_{1,-q}$ is a q-measure on ${\mathbb{Z}}_p$. Hence, we obtain the following theorem.

Theorem 3 Let ${\tilde{\mu }}_{-q}$ be a strongly fermionic weighted p-adic q-measure on ${\mathbb{Z}}_p$, and assume that the fermionic weighted Radon-Nikodym derivative $f_{{\tilde{\mu }}_{-q}}$ on ${\mathbb{Z}}_p$ is a continuous function on ${\mathbb{Z}}_p$. Suppose that ${\tilde{\mu }}_{1,-q}$ is the strongly fermionic weighted p-adic q-measure associated to $f_{{\tilde{\mu }}_{-q}}$. Then there exists a q-measure ${\tilde{\mu }}_{2,-q}$ on ${\mathbb{Z}}_p$ such that

$${\tilde{\mu }}_{-q}={\tilde{\mu }}_{1,-q}+{\tilde{\mu }}_{2,-q}.$$

(2.30)