<figure><img src="image_1.png" /></figure>

BLAISE PASCAL

HAMADOUN MAÏGA

Integrable functions for the Bernoulli measures of rank 1

Volume 17, n° 2 (2010), p. 341-356.

<http://ambp.cedram.org/item?id=AMBP_2010__17_2_341_0>

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Publication éditée par le laboratoire de mathématiques
de l'université Blaise-Pascal, UMR 6620 du CNRS
Clermont-Ferrand — France
# Integrable functions for the Bernoulli measures of rank 1

HAMADOUN MAÏGA

## Abstract

In this paper, following the p-adic integration theory worked out by A. F. Monna and T. A. Springer [4, 5] and generalized by A. C. M. van Rooij and W. H. Schikhof [6, 7] for the spaces which are not σ-compacts, we study the class of integrable p-adic functions with respect to Bernoulli measures of rank 1. Among these measures, we characterize those which are invertible and we give their inverse in the form of series.

## 1. Preliminaries.

In what follows, we denote by $p$ a prime number, $\mathbb{Q}$ the field of rational numbers provided with the $p$-adic absolute value, $\mathbb{Q}_p$ the field of $p$-adic numbers that is the completion of $\mathbb{Q}$ for the $p$-adic absolute value and by $\mathbb{Z}_p$ the ring of $p$-adic integers. We denote by $v_p$ the normalized valuation of $\mathbb{Q}_p$.

Let $X$ be a totally discontinuous compact space and $\Omega(X)$ the Boolean algebra of closed and open subsets of $X$. If $U$ belongs to $\Omega(X)$, one denotes by $\chi_U$ the characteristic function of $U$ which is a continuous function. For $K$ a complete ultrametric valued field, $C(X, K)$ is the Banach algebra of the continuous functions from $X$ into $K$ provided with the norm of uniform convergence, $\|f\|_{\infty} = \sup_{x \in X} |f(x)|$.

**Definition 1.1.** A measure on $X$ is an additive map $\mu: \Omega(X) \to K$ such that

$$ \|\mu\| = \sup_{V \in \Omega(X)} |\mu(V)| < +\infty. $$

One denotes by $M(X, K)$ the space of measures on $X$. Provided with the norm $\|\mu\| = \sup_{U \in \Omega(X)} |\mu(U)|$, it is an ultrametric $K$-Banach space.

<footnote>Keywords: integrable functions, Bernoulli measures of rank 1, invertible measures.</footnote>

<footnote>Math. classification: 46S10.</footnote>
Let $\mu$ be a measure on $X$; for any locally constant function

$$f = \sum_{j=1}^{n} \lambda_{j} \chi_{U_{j}},$$

putting $\varphi_{\mu}(f) = \sum_{j=1}^{n} \lambda_{j}\mu(U_{j})$, one defines on the space $Loc(X, K)$ of the locally constant functions a continuous linear form such that

$$|\varphi_{\mu}(f)| \leq \|\mu\| \|f\|_{\infty},$$

then

$$\|\varphi_{\mu}\| = \sup_{f \neq 0} \frac{|\varphi_{\mu}(f)|}{\|f\|_{\infty}} \leq \|\mu\|.$$

The linear form $\varphi_{\mu}$ on $Loc(X, K)$ associated to $\mu$ being continuous for the uniform norm on $Loc(X, K)$ and since this space is a dense subspace of $C(X, K)$, one sees that $\varphi_{\mu}$ extends to an unique continuous linear form on $C(X, K)$ with the same norm and also noted $\varphi_{\mu}$.

On the other hand, if $\varphi$ is a continuous linear form on the Banach space $C(X, K)$, by setting for any closed and open subset $U$ of $X$: $\mu_{\varphi}(U) = \varphi(\chi_{U})$, one defines a measure $\mu_{\varphi}$ on $X$ such that $\|\mu_{\varphi}\| \leq \|\varphi\|$.

Therefore, a measure $\mu = \mu_{\varphi}$ on $X$ which corresponds to some continuous linear form $\varphi$ on $C(X, K)$ is such that $\varphi = \varphi_{\mu}$ and $\|\varphi_{\mu}\| = \|\mu\|$. Hence one sees that $M(X, K)$ is isometrically isomorphic to the dual Banach space $C(X, K)'$ of $C(X, K)$.

Let $\mu$ be a measure on $X$ and $\varphi_{\mu}$ the continuous linear form associated to $\mu$. One defines an ultrametric seminorm on $C(X, K)$ by setting, for $f \in C(X, K)$:

$$\|f\|_{\mu} = \sup_{g \in C(X, K), g \neq 0} \frac{|\mu(fg)|}{\|g\|_{\infty}},$$

where $\mu(f) = \varphi_{\mu}(f)$ for $f \in C(X, K)$.

Let us remind some fundamental notions on p-adic integration theory.

**Theorem 1.2** (Schikhof). For any $\mu \in M(X, K)$, there exists a unique upper semicontinuous function $N_{\mu}: X \to [0, \infty)$ such that

$$\|f\|_{\mu} = \sup_{x \in X} |f(x)| N_{\mu}(x).$$

The function $N_{\mu}$ is given by the formula

$$N_{\mu}(x) = \inf_{U \in \Omega(X), x \in U} \|\chi_{U}\|_{\mu}.$$
## $\mu_{1,\alpha}$-INTEGRABLE FUNCTIONS

*Proof.* See [7, page 278] or [6, Lemma 7.2] for a proof of the theorem. □

For any closed and open subset $U \in \Omega(X)$

$$ \|\chi_U\|_{\mu} = \sup_{V \subset U, V \in \Omega(X)} |\mu(V)|. $$

This relation is important for many computations which will follow.

If $\mu$ is a measure on $X$ and $f: X \to K$ a locally constant function, one sets $\varphi_{\mu}(f) = \int_X f(x)d\mu(x)$, called the integral of $f$ with respect to $\mu$.

**Definition 1.3** ($\mu$-integrable functions). Let $\mu$ be a measure on $X$ and $f: X \to K$ a function; one puts

$$ \|f\|_s = \sup_{x \in X} |f(x)| N_{\mu}(x). $$

One says that:

* $f$ is $\mu$-negligible if $\|f\|_s = 0$ and a subset $U$ of $X$ is $\mu$-negligible if $\|\chi_U\|_s = 0$.

* $f$ is $\mu$-integrable if there exists a sequence $(f_n)_n$ of locally constant functions such that $\lim_{n \to +\infty} \|f - f_n\|_s = 0$.

One sets $\int_X f(x)d\mu(x) = \lim_{n \to +\infty} \int_X f_n(x)d\mu(x)$, which is seen to be independent of the sequence $(f_n)_n$.

For $x \in X$ and $\mu$ be a measure on $X$, one has $N_{\mu}(x) \le \|\mu\|$ and one can show that $\|f\|_s = \|f\|_{\mu}$ for any continuous functions $f: X \to K$.

In the sequel, one denotes by $\mathcal{L}^1(X, \mu)$ the spaces of $\mu$-integrable functions and by $L^1(X, \mu)$ the quotient space $\mathcal{L}^1(X, \mu)/\Re$, where $\Re$ is the equivalence relation defined by $f \Re g$ if $f - g$ is $\mu$-negligible.

For any continuous function $f: X \to K$, one has $\|f\|_s \le \|\mu\|\|f\|_{\infty}$.

Since the space of locally constant functions is uniformly dense in the space of continuous functions, one sees that any continuous function is $\mu$-integrable, in other words : $C(X, K) \subseteq \mathcal{L}^1(X, K)$.

Furthermore if $f \in C(X, K)$, one has $\int_X f(x)d\mu(x) = \varphi_{\mu}(f)$.

## 2. Integrable functions for the Bernoulli measures of rank 1.

We assume now that the complete valued field $K$ is a valued extension of $\mathbb{Q}_p$ and we let $\alpha$ be a $p$-adic unit. Let us remind that the Bernoulli
polynomials $(B_n(x))_{n \ge 0}$ are defined by

$$ \frac{te^{xt}}{e^t - 1} = \sum_{k \ge 0} B_k(x) \frac{t^k}{k!} $$

**Definition 2.1** (Koblitz [3], Proposition, page 35). Let $k \ge 1$ be a fixed integer and $B_k(x)$ be the $k$-th Bernoulli polynomial. For any integer $n \ge 1$ and $a \in \{0, 1, \dots, p^n - 1\}$, put

$$ \mu_k(a + p^n \mathbb{Z}_p) = p^{n(k-1)} B_k\left(\frac{a}{p^n}\right). $$

If $U = \cup_{i=1}^N (a_i + p^{n_i} \mathbb{Z}_p)$ is a partition of the closed and open subset $U$ of $\mathbb{Z}_p$, setting $\mu_k(U) = \sum_{i=1}^N \mu_k(a_i + p^{n_i} \mathbb{Z}_p)$, one can prove, with properties of Bernoulli polynomials, that this sum is independent of any such partition of $U$ and one obtains an additive map $\mu_k: \Omega(\mathbb{Z}_p) \to K$ called the *Bernoulli distribution* of rank $k$.

**Definition 2.2** (B. Mazur). Let $k \ge 1$ be a fixed integer and $\alpha$ be a $p$-adic unit. The *Bernoulli measure* of rank $k$ normalized by $\alpha$ is the measure defined by setting for any closed and open set $U \in \Omega(\mathbb{Z}_p)$

$$ \mu_{k,\alpha}(U) = \mu_k(U) - \alpha^{-k} \mu_k(\alpha U). \qquad (2.1) $$

Let $a$ be a $p$-adic integer, whose Hensel expansion is $a = \sum_{i \ge 0} a_i p^i$. For an integer $n \ge 1$, one puts

$$ (a)_n = \sum_{i<n} a_i p^i \quad \text{and} \quad [a]_n = \sum_{i \ge 0} a_{n+i} p^i. $$

One has then

$$ [a]_n = \frac{a}{p^n} - \frac{(a)_n}{p^n} \in \mathbb{Z}_p \text{ and } [a\alpha]_n = \frac{a\alpha}{p^n} - \frac{(a\alpha)_n}{p^n} \in \mathbb{Z}_p. $$

Setting $U = a + p^n \mathbb{Z}_p$ for any integer $a \in \{0, 1, \dots, p^n - 1\}$ and $k = 1$ in the relation (2.1), one has $\mu_{1,\alpha}(a + p^n \mathbb{Z}_p) = B_1(\frac{a}{p^n}) - \alpha^{-1} B_1(\frac{(a\alpha)_n}{p^n})$.

As $B_1(x) = x - 1/2$, one obtains

$$ \mu_{1,\alpha}(a + p^n \mathbb{Z}_p) = \left(\frac{a}{p^n} - \frac{1}{2}\right) - \alpha^{-1}\left(\frac{a\alpha}{p^n} - [a\alpha]_n - \frac{1}{2}\right). $$

Thus, for all integers $n \ge 1$ and $a \in \{0, 1, \dots, p^n - 1\}$,

$$ \mu_{1,\alpha}(a + p^n \mathbb{Z}_p) = \frac{1}{2\alpha}(1 - \alpha + 2[a\alpha]_n). \qquad (2.2) $$
## $\mu_{1,\alpha}$-INTEGRABLE FUNCTIONS

**Proposition 2.3.** The measure $\mu_{1,-1}$ is equal to $-\delta_0$, where $\delta_0$ is the Dirac measure at 0. The space of $\mu_{1,-1}$-integrable functions is equal to the space of all functions $f: \mathbb{Z}_p \to K$.

*Proof.* Let $n \ge 1$ be an integer, and $a$ be an integer such that $0 \le a \le p^n - 1$; according to the relation (2.2), one has: $\mu_{1,-1}(a + p^n\mathbb{Z}_p) = -1 - [-a]_n$.

* For $a = 0$, it follows that $\mu_{1,-1}(p^n\mathbb{Z}_p) = -1$.

* For $1 \le a \le p^n - 1$, one has $-(p^n - 1) \le -a \le -1$ and
  $1 \le p^n - a \le p^n - 1$.

As $-1 = \sum_{i \ge 0} (p-1)p^i$, one has
$$ -a = (p^n - a) - p^n = (p^n - a) + p^n \sum_{i \ge 0} (p-1)p^i. $$

Hence, one has $[-a]_n = \sum_{i \ge 0} (p-1)p^i = -1$ and $\mu_{1,-1}(a + p^n\mathbb{Z}_p) = 0$.

Let $\delta_0$ be the Dirac measure at 0. It is readily seen that $\mu_{1,-1} = -\delta_0$ and $L^1(\mathbb{Z}_p, \mu_{1,-1})$ is algebraically isomorphic to $K$. $\square$

We now assume that $\alpha$ is a $p$-adic unit different from 1 and of $-1$ and we set $\gamma_\alpha = \inf_{x \in \mathbb{Z}_p} N_{\mu_{1,\alpha}}(x)$.

Let $j \ge 1$ be an integer; for any integer $a \in \{0, 1, \dots, p^j - 1\}$, one has

$$ |\mu_{1,\alpha}(a + p^j \mathbb{Z}_p)| = \left| \frac{1}{2\alpha} (1 - \alpha + 2[a\alpha]_j) \right| \le \max\left( |1 - \alpha|, |2[a\alpha]_j| \right) \le 1. $$

Let us remind that any closed and open subset $V$ of $\mathbb{Z}_p$ can be written as disjoint union $V = \bigsqcup_{k=1}^m (a_k + p^{jk}\mathbb{Z}_p)$. Hence, one has

$$ |\mu_{1,\alpha}(V)| \le \max_{1 \le k \le m} |\mu_{1,\alpha}(a_k + p^{jk}\mathbb{Z}_p)| \le 1. $$

Thus, for all integer $n \ge 1$ and $a$ such that $0 \le a \le p^n - 1$, one has

$$ \|\chi_{a+p^n\mathbb{Z}_p}\|_{\mu_{1,\alpha}} = \sup_{V \subset a+p^n\mathbb{Z}_p} |\mu_{1,\alpha}(V)| \le 1 $$

Moreover, we have $N_{\mu_{1,\alpha}}(x) \le \|\mu_{1,\alpha}\| \le 1$, for any $p$-adic integer $x$.

**Lemma 2.4.** Let $\alpha = 1 + bp^r$ be a principal unit of the ring of $p$-adic integer, different from 1, with $r = v_p(\alpha - 1) \ge 1$. For any $p$-adic integer $x$, one has

* $N_{\mu_{1,\alpha}}(x) \ge \frac{1}{p^r}$, if $p$ is odd;
• $N_{\mu_{1,\alpha}}(x) \ge \frac{1}{2^{r-1}}$, if $p=2$ and $r \ge 2$.

Therefore $\gamma_{\alpha} \ge \frac{1}{p^r}$ if $p \ne 2$ and $\gamma_{\alpha} \ge \frac{1}{2^{r-1}}$ if $p=2$ and $r \ge 2$.

*Proof.* Let us remind that

$$ \mu_{1,1+bp^r}(a + p^n\mathbb{Z}_p) = \frac{1}{\alpha}([a\alpha]_n - \frac{1}{2}bp^r), $$

where $n$ and $a$ be an integers such that $n \ge 1$ and $a \in \{0, 1, \dots, p^n - 1\}$, and where $r = v_p(\alpha - 1) \ge 1$. One has two cases :

**First case :** $p$ odd.

• If $a=0$, one has $|\mu_{1,\alpha}(p^n\mathbb{Z}_p)| = \frac{1}{p^r}$; it follows that $\|\chi_{p^n\mathbb{Z}_p}\|_{\mu_{1,\alpha}} \ge \frac{1}{p^r}$.

• Now, assume that $1 \le a \le p^n - 1$;

(1) If $|[a\alpha]|_n < \frac{1}{p^r}$, one has $|\mu_{1,1+bp^r}(a + p^n\mathbb{Z}_p)| = |\frac{1}{2}bp^r| = \frac{1}{p^r}$;

(2) If $|[a\alpha]|_n > \frac{1}{p^r}$, one has $|\mu_{1,1+bp^r}(a + p^n\mathbb{Z}_p)| = |[a\alpha]|_n > \frac{1}{p^r}$. In these two cases, one obtains $\|\chi_{a+p^n\mathbb{Z}_p}\|_{\mu_{1,\alpha}} \ge \frac{1}{p^r}$.

(3) If $|[a\alpha]|_n = \frac{1}{p^r}$, consider $c_n + c_{n+1}p + \dots + c_{n+r}p^r + \dots$ the Hensel expansion of $[a\alpha]_n$. One then has $c_n = c_{n+1} = \dots = c_{n+r-1} = 0$ and $c_{n+r} \ne 0$. It follows that $[a\alpha]_{n+1} = c_{n+r}p^{r-1} + c_{n+r+1}p^r + \dots$; since $|2[a\alpha]_{n+1}| = |[a\alpha]_{n+1}|$, one has

$$ |\mu_{1,\alpha}(a + p^{n+1}\mathbb{Z}_p)| = \frac{1}{p^{r-1}} \ge \frac{1}{p^r} $$

$$ \text{and } \|\chi_{a+p^n\mathbb{Z}_p}\|_{\mu_{1,\alpha}} \ge \frac{1}{p^r}. $$

Let $V_x$ be an open and closed neighborhood of $x$. There exists an integer $j_0 \ge 1$ such that $x + p^{j_0}\mathbb{Z}_p \subset V_x$. Thus, one has $\|\chi_{x+p^{j_0}\mathbb{Z}_p}\|_{\mu_{1,\alpha}} \le \|\chi_{V_x}\|_{\mu_{1,\alpha}}$.

It follows that $\|\chi_{V_x}\|_{\mu_{1,\alpha}} \ge \frac{1}{p^r}$. Taking infimum, one obtains

$$ N_{\mu_{1,\alpha}}(x) \ge \frac{1}{p^r}. $$

**Second case :** $p=2$ and $r \ge 2$. Putting $\alpha = 1 + 2^r b$, one has

$$ \mu_{1,\alpha}(a + 2^n\mathbb{Z}_2) = \frac{1}{\alpha}([a\alpha]_n - 2^{r-1}b). $$
* If $a = 0$, for any integer $n \ge 1$, one has $|\mu_{1,\alpha}(2^n\mathbb{Z}_2)| = \frac{1}{2^{r-1}}$.
  It follows that $\|\chi_{2^n\mathbb{Z}_2}\|_{\mu_{1,\alpha}} \ge \frac{1}{2^{r-1}}$.

* Let us suppose that $1 \le a \le 2^n - 1$.

(1) If $|[a\alpha]|_n < \frac{1}{2^{r-1}}$, one has $|\mu_{1,\alpha}(a + 2^n\mathbb{Z}_2)| = \frac{1}{2^{r-1}}$;

(2) If $|[a\alpha]|_n > \frac{1}{2^{r-1}}$, one has $|\mu_{1,\alpha}(a + 2^n\mathbb{Z}_2)| = |[a\alpha]|_n > \frac{1}{2^{r-1}}$.
In these two cases, one obtains $\|\chi_{a+2^n\mathbb{Z}_2}\|_{\mu_{1,\alpha}} \ge \frac{1}{2^{r-1}}$.

(3) If $|[a\alpha]|_n = \frac{1}{2^{r-1}}$, as in the First case (3), one shows that
$\|\chi_{a+2^n\mathbb{Z}_2}\|_{\mu_{1,\alpha}} \ge \frac{1}{2^{r-1}}$.

Let $x \in \mathbb{Z}_2$; one shows as in the First case that $N_{\mu_{1,\alpha}}(x) \ge \frac{1}{2^{r-1}}$ and again that $\gamma_\alpha \ge \frac{1}{2^{r-1}}$.

□

**Lemma 2.5.** Let $p$ be an odd prime number, $\alpha = \alpha_0+bp^r$ be a $p$-adic unit, where $\alpha_0$ is an integer such that $2 \le \alpha_0 \le p-1$ and $r = v_p(\alpha - \alpha_0) \ge 2$. One has $N_{\mu_{1,\alpha}}(x) \ge \frac{1}{p^r}$ for any $p$-adic integer $x$ and $\gamma_\alpha \ge \frac{1}{p^r}$.

*Proof.* Let $p$ be an odd prime number and $\alpha = \alpha_0 + bp^r$ be a $p$-adic unit, where $\alpha_0$ is an integer such that $2 \le \alpha_0 \le p-1$ and $r = v_p(\alpha - \alpha_0) \ge 2$. Let us remind that, for all integers $n \ge 1$ and $a$ such that $0 \le a \le p^n - 1$, one has

$$ \mu_{1,\alpha}(a + p^n\mathbb{Z}_p) = \frac{1}{\alpha}([a\alpha]_n + \frac{1-\alpha}{2}) = \frac{1}{\alpha}([a\alpha]_n - \frac{\alpha_0 - 1}{2}) - \frac{1}{2}bp^r]. $$

* If $a = 0$, one has $|\mu_{1,\alpha}(p^n\mathbb{Z}_p)| = |\frac{1-\alpha}{2\alpha}| = 1$;

* Let us suppose that $a \in \{1, 2, \dots, p^n - 1\}$.

(1) If $|[a\alpha]_n - \frac{\alpha_0 - 1}{2}| < \frac{1}{p^r}$, one has $|\mu_{1,\alpha}(a + p^n\mathbb{Z}_p)| = |\frac{1}{2}bp^r| = \frac{1}{p^r}$.

(2) If $|[a\alpha]_n - \frac{\alpha_0 - 1}{2}| > \frac{1}{p^r}$, one has $|\mu_{1,\alpha}(a + p^n\mathbb{Z}_p)| > \frac{1}{p^r}$.
In these two cases, one obtains $\|\chi_{a+p^n\mathbb{Z}_p}\|_{\mu_{1,\alpha}} \ge \frac{1}{p^r}$.

(3) If $|[a\alpha]_n - \frac{\alpha_0 - 1}{2}| = \frac{1}{p^r}$, let $c_n + c_{n+1}p + \dots$ be the Hensel expansion of $[a\alpha]_n$; there are two cases according to the parity of $\alpha_0$:

**First case :** $\alpha_0$ odd.

One has $c_n = \frac{\alpha_0 - 1}{2}$, $c_{n+1} = \dots = c_{n+r-1} = 0$ and $c_{n+r} \ne 0$.
Hence $[a\alpha]_{n+1} = c_{n+r}p^{r-1} + c_{n+r+1}p^r + \dots$. It follows that

$$|[a\alpha]_{n+1} - \frac{\alpha_0 - 1}{2}| = |\frac{\alpha_0 - 1}{2}| = 1$$

and $|\mu_{1,\alpha}(a + p^{n+1}\mathbb{Z}_p)| = 1$.

**Second case :** $\alpha_0$ even.

The Hensel expansion of $\frac{\alpha_0-1}{2}$ is

$$\frac{\alpha_0 - 1}{2} = \frac{p + \alpha_0 - 1}{2} + \sum_{i \ge 1} \frac{p - 1}{2} p^i.$$

In this case, one has : $c_n = \frac{p+\alpha_0-1}{2}$ and for $j \in \{n+1, \dots, n+r-1\}$, $c_j = \frac{p-1}{2}$. Hence,

$$[a\alpha]_{n+1} = \sum_{i=0}^{r-2} \frac{p-1}{2} p^i + \sum_{i \ge r-1} c_{n+i+1} p^i.$$

Therefore

$$[a\alpha]_{n+1} - \frac{\alpha_0 - 1}{2} = -\frac{\alpha_0}{2} + \sum_{i \ge r-1} \left(c_{n+i+1} - \frac{p-1}{2}\right) p^i.$$

Thus, one has $|[a\alpha]_{n+1} - \frac{\alpha_0-1}{2}| = | - \frac{\alpha_0}{2}| = 1$ and

$$|\mu_{1,\alpha}(a + p^{n+1}\mathbb{Z}_p)| = 1.$$

Finally, in these two cases, we have proved that

$$\|\chi_{a+p^n\mathbb{Z}_p}\|_{\mu_{1,\alpha}} \ge \frac{1}{p^r}.$$

As in the proof of Lemma 2.4, one proves that $N_{\mu_{1,\alpha}}(x) \ge p^{-r}$, for any $p$-adic integer $x$ and $\gamma_\alpha = \inf_{x \in \mathbb{Z}_p} N_{\mu_{1,\alpha}}(x) \ge p^{-r}$. $\square$

**Lemma 2.6.** • Let $p$ be an odd prime number and $\alpha$ be an integer $\ge 2$ which is a $p$-adic unit not congruent to $1$ modulo $p$, then $\gamma_\alpha = 1$.

* Let $\alpha$ be a negative integer $< -1$ which is a $p$-adic unit; one has then

$$\gamma_{\alpha} \ge \min \left( \left| \frac{1-\alpha}{2} \right|, \left| \frac{1+\alpha}{2} \right| \right) > 0.$$
$\mu_{1,\alpha}$-INTEGRABLE FUNCTIONS

*Proof.* Let us remind that $\mu_{1,\alpha}(a + p^n\mathbb{Z}_p) = \frac{1}{2\alpha}(1 - \alpha + 2[a\alpha]_n)$, for all integers $n \ge 1$ and $a \in \{0, 1, \dots, p^n - 1\}$.

• Let $p$ be an odd prime number and $\alpha \ge 2$ be an integer which is a $p$-adic unit such that $\alpha \ne 1 \pmod p$ and $n \ge 1$ be a fixed integer; let us consider an integer $a$ such that $0 \le a \le p^n - 1$.

If $a=0$, one has $\|\chi_{p^n\mathbb{Z}_p}\|_{\mu_{1,\alpha}} \ge |\mu_{1,\alpha}(p^n\mathbb{Z}_p)| = |\alpha - 1| = 1$.

Now, let us suppose that $1 \le a \le p^n - 1$ and let us consider an integer $j$ such that $p^j \ge \alpha p^n - \alpha + 1$; one has $\alpha \le a\alpha \le \alpha p^n - \alpha < p^j$. It follows that $(a\alpha)_j = a\alpha$ and $[a\alpha]_j = 0$. In this case, one has $|\mu_{1,\alpha}(a + p^j\mathbb{Z}_p)| = 1$. Hence, one has $\|\chi_{a+p^n\mathbb{Z}_p}\|_{\mu_{1,\alpha}} \ge 1$, for all integers $n$ and $a$ such that $n \ge 1$ and $a \in \{0, 1, \dots, p^n - 1\}$.

Thus, as in the proof of Lemma 2.4, we have $N_{\mu_{1,\alpha}}(x) \ge 1$, for any $p$-adic integer $x$. Since $N_{\mu_{1,\alpha}}(x) \le 1$, for any $p$-adic integer $x$, the function $N_{\mu_{1,\alpha}}$ is constant and $N_{\mu_{1,\alpha}}(x) = 1 = \gamma_{\alpha}$.

• Let $\alpha$ be a negative integer $< -1$ which is a $p$-adic unit, $n \ge 1$ be an integer and $a \in \{1, 2, \dots, p^n - 1\}$. One obtains a strictly positive integer while setting $m = -a\alpha$; let us denote by $s(m)$ the highest power of $p$ in the Hensel expansion of $m$. One has two cases:

**First case :** $m = p^{s(m)}$. One has $a\alpha = -m = p^{s(m)} \sum_{k \ge 0} (p-1)p^k$. Thus $[a\alpha]_j = \sum_{i \ge 0} (p-1)p^i = -1$ and $\mu_{1,\alpha}(a + p^j\mathbb{Z}_p) = -\frac{\alpha+1}{2\alpha}$, for any integer $j > \max(s(m), n)$. It follows that

$$|\mu_{1,\alpha}(a + p^j\mathbb{Z}_p)| = \left|\frac{\alpha + 1}{2\alpha}\right| = \left|\frac{\alpha + 1}{2}\right| \le \|\chi_{a+p^n\mathbb{Z}_p}\|_{\mu_{1,\alpha}}.$$

**Second case :** $m \ne p^{s(m)}$. One has $-m = (p^{s(m)+1}-m)-p^{s(m)+1}$; the Hensel expansion of $a\alpha = -m$ is given by

$$a\alpha = \sum_{\ell=0}^{s(m)} \beta_{\ell} p^{\ell} + \sum_{j \ge 0} (p-1) p^{s(m)+1+j}.$$

Thus, for any integer $j > \max(s(m) + 1, n)$, one has $[a\alpha]_j = \sum_{i \ge 0} (p-1)p^i = -1$ and

$$|\mu_{1,\alpha}(a + p^j\mathbb{Z}_p)| = \left|\frac{\alpha + 1}{2}\right| \le \|\chi_{a+p^n\mathbb{Z}_p}\|_{\mu_{1,\alpha}}.$$
On other hand, one has

$$ \|\chi_{p^n\mathbb{Z}_p}\|_{\mu_{1,\alpha}} \geq |\mu_{1,\alpha}(p^n\mathbb{Z}_p)| = \left|\frac{\alpha - 1}{2}\right|. $$

It follows that $\|\chi_{a+p^n\mathbb{Z}_p}\|_{\mu_{1,\alpha}} \geq \min(|\frac{\alpha+1}{2}|, |\frac{\alpha-1}{2}|)$, for any integer $n \geq 1$ and any integer $a \in \{0, 1, \dots, p^n - 1\}$. One concludes that $\gamma_{\alpha} > 0$.

□

**Theorem 2.7.** Let $\alpha$ be a $p$-adic unit of one of the following forms :

* $\alpha = 1 + bp^r$, where $r = v_p(\alpha - 1)$ is such that $r \geq 1$ if $p \neq 2$ and $r \geq 2$ if $p = 2$;

* $\alpha = \alpha_0 + bp^r$, if $p \neq 2$, $\alpha_0 \in \{2, \dots, p-1\}$ and $r = v_p(\alpha - \alpha_0) \geq 2$;

* $\alpha \geq 2$ is an integer such that $\alpha \neq 1 \pmod{p}$ (with $p$ odd).

* $\alpha$ is a negative integer different from $-1$.

The space $\mathcal{L}^1(\mathbb{Z}_p, \mu_{1,\alpha})$ of $\mu_{1,\alpha}$-integrable functions is equal to the space of continuous functions $C(\mathbb{Z}_p, K)$.

Furthermore, one has $L^1(\mathbb{Z}_p, \mu_{1,\alpha}) = C(\mathbb{Z}_p, K)$.

*Proof.* Let us suppose that the conditions on $\alpha$ (of Theorem 2.7) are satisfied. According to Lemmas 2.4, 2.5 and 2.6, one has $\gamma_{\alpha} > 0$. Hence, one has $\gamma_{\alpha} \leq N_{\mu_{1,\alpha}}(x) \leq 1$ for any $p$-adic integers $x$. Thus, for any function $f: \mathbb{Z}_p \to K$, one has

$$ \gamma_{\alpha} \|f\|_{\infty} \leq \|f\|_{s} \leq \|f\|_{\infty}. $$

Let us assume that $f: \mathbb{Z}_p \to K$ is a $\mu_{1,\alpha}$-integrable function. There exists a sequence $(f_n)_{n \geq 0}$ of locally constant functions such that $\lim_{n \to +\infty} \|f - f_n\|_s = 0$. Since $\gamma_{\alpha} \|f - f_n\|_{\infty} \leq \|f - f_n\|_s$, $f$ is a uniform limit of continuous functions. Hence $f$ is continuous. It follows that $\mathcal{L}^1(\mathbb{Z}_p, \mu_{1,\alpha}) = C(\mathbb{Z}_p, K)$.

Moreover the null function is the only $\mu_{1,\alpha}$-negligible function and one has $L^1(\mathbb{Z}_p, \mu_{1,\alpha}) = C(\mathbb{Z}_p, K)$. □

**Remark 2.8.** It remains to characterize the $\mu_{1,\alpha}$-integrable functions, where $\alpha = \alpha_0 + bp$ is not an integer, $\alpha_0 \in \{2, \dots, p-1\}$ and $v_p(\alpha - \alpha_0) = 1$ for $p \neq 2$ and $\alpha = 1 + 2b$ not a negative integer $\leq -1$ with $v_2(\alpha - 1) = 1$ for $p = 2$.
### 3. Inversibility of measures $\mu_{1,\alpha}$.

In what follows, if $n \ge 1$ is an integer, we denote by $s(n)$ the highest power of $p$ in the Hensel expansion of $n$.

**Lemma 3.1.** Let $p$ be a prime number and $\alpha = 1 + bp^r$ be a principal unit of the ring of p-adic integer different of 1 (with $r = v_p(\alpha - 1) \ge 1$).

There exists an integer $n \ge 1$ such that $|[n\alpha]_{s(n)+1}| = 1$.

*Proof.* Let $\alpha = 1 + bp^r$ be a principal unit of the ring of p-adic integer different from 1 (with $r = v_p(\alpha - 1) \ge 1$) and $m$ be an integer such that $m \ge r$. Let us consider the positive integer $n = p^{m-r+1}(1+p+\cdots+p^{r-1})$; one has $s(n) = m$ and $n = p^{s(n)-r+1} + \cdots + p^{s(n)}$. Hence, one has

$$ 
\begin{aligned}
n\alpha &= [p^{s(n)-r+1} + \cdots + p^{s(n)}](1 + bp^r) \\
&= p^{s(n)-r+1} + \cdots + p^{s(n)} + bp^{s(n)+1}(1 + p + \cdots + p^{r-1})
\end{aligned}
$$

It follows that $[n\alpha]_{s(n)+1} = b(1+p+\cdots+p^{r-1})$ and $|[n\alpha]_{s(n)+1}| = 1$. $\square$

**Definition 3.2.** Let $n$ be an integer $\ge 0$, and $x$ be a p-adic integer. The integer $n$ is called an initial part of $x$, and one notes $n\triangleleft x$ if $|x-n| < \frac{1}{n}$.

M. van der Put could showed that the sequence of functions $(e_n)_n$ defined by $e_n(x) = \begin{cases} 1 & \text{if } n \triangleleft x \\ 0 & \text{otherwise} \end{cases}$. is an orthonormal base of $C(\mathbb{Z}_p, K)$; $(e_n)_n$ is called the van der Put base.

**Theorem 3.3.** Let $p$ be a prime number and $\alpha \ne 1$ be a p-adic unit. Then $\|\mu_{1,\alpha}\| = 1$.

*Proof.* • If $p$ be an odd prime number and $\alpha$ be a p-adic unit of the form $\alpha = \alpha_0 + bp^r$, where $\alpha_0 \in \{2, 3, \cdots, p-1\}$ and $r = v_p(\alpha - \alpha_0) \ge 1$, one has $1 = |\frac{1-\alpha}{2\alpha}| = |\langle\mu_{1,\alpha}, e_0\rangle| \le \|\mu_{1,\alpha}\| \le 1$. Hence, one has $\|\mu_{1,\alpha}\| = 1$.

• If $\alpha = 1 + bp^r$ is a principal unit of the ring of p-adic integers, different from 1, one has two cases :

(1) $p \ne 2$ or $r \ge 2$. According to Lemma 3.1, there exists an integer $n_0 \ge 1$ such that $|[n_0\alpha]_{s(n_0)+1}| = 1$. In this case, one has $1 = |\langle\mu_{1,\alpha}, e_{n_0}\rangle| \le \|\mu_{1,\alpha}\| \le 1$. It follows that $\|\mu_{1,\alpha}\| = 1$.
(2) $r=1$ and $p=2$.

One has $\alpha = 1 + 2b$ and $\langle \mu_{1,1+2b}, e_0 \rangle = \frac{-2b}{2(1+2b)} = \frac{-b}{\alpha}$.

Hence,

$$1 = \left| \frac{-b}{\alpha} \right| = \left| \langle \mu_{1,1+2b}, e_0 \rangle \right| \le \| \mu_{1,1+2b} \| \le 1$$

and $\|\mu_{1,1+2b}\| = 1$.

• If $\alpha$ is a $p$-adic unit such that $2 \le \alpha \le p-1$ (with $p$ odd), one has

$$1 = \left| \frac{1-\alpha}{2\alpha} \right| = \left| \langle \mu_{1,\alpha}, e_0 \rangle \right| \le \| \mu_{1,\alpha} \| \le 1.$$

It follows that $\|\mu_{1,\alpha}\| = 1$.

<figure><img src="image_2.png" /></figure>

Let $\delta_a$ be the Dirac measure associated to the $p$-adic integer $a$. Let us put $\omega = \delta_1 - \delta_0$. It is known that any measure $\mu \in M(\mathbb{Z}_p, K)$ can be written as a pointwise convergent series $\mu = \sum_{n \ge 0} \langle \mu, Q_n \rangle Q'_n$, where $(Q_n)_{n \ge 0}$ is an orthonormal base of $C(\mathbb{Z}_p, K)$ called the Mahler basis, defined by $Q_n(x) = \binom{x}{n}$ and $(Q'_n)_{n \ge 0}$ is the dual family of $(Q_n)_{n \ge 0}$ defined by $\langle Q'_n, Q_m \rangle = \delta_{nm}$. The convolution product $Q'_n \star Q'_m$ gives $Q'_{n+m}$; one deduces that $Q'_n = Q'_1$. As $Q'_1 = \omega$, one has $Q'_n = \omega^n$; it follows that $\mu = \sum_{n \ge 0} \langle \mu, Q_n \rangle \omega^n$. Hence the measure $\mu$ corresponds to the formal power series of bounded coefficients $S_\mu = \sum_{n \ge 0} \langle \mu, Q_n \rangle X^n$. Therefore, the algebra $M(\mathbb{Z}_p, K)$, provided with the convolution product, is isometrically isomorphic to the algebra $K\langle X \rangle$ of bounded formal power series with bounded coefficients, the norm being the supremum of the coefficients.

Let us remind (see for instance [2] or [1]) that an element $S$ of the Banach algebra $K\langle X \rangle$ is invertible if and only if $\|S\| = |S(0)| \ne 0$.

**Theorem 3.4.** Let $p$ be a prime number, and $\alpha$ be a $p$-adic unit.

The measure $\mu_{1,\alpha}$ is invertible for the convolution product if and only if $\alpha \ne 1 \pmod p$ if $p$ is odd (resp. $\alpha \ne 1 \pmod 4$ for $p=2$).

Moreover its inverse $\nu_\alpha$ is given by the formula

$$\nu_\alpha = \sum_{n \ge 0} d_n(\alpha) \omega^n,$$
where $d_0(\alpha) = \frac{2\alpha}{1-\alpha}$, $d_1(\alpha) = \frac{1+\alpha}{3(1-\alpha)}$ and for $n \ge 2$:

$$d_n(\alpha) = \alpha^n \left(\frac{2\alpha}{1-\alpha}\right)^{n+1} \sum_{\substack{1 \le j \le n \\ i_1 + \dots + i_j = n \\ i_1, \dots, i_j \in \{1, \dots, n\}}} (-1)^j \binom{\alpha^{-1}}{i_1 + 2} \cdots \binom{\alpha^{-1}}{i_j + 2}.$$

*Proof.* Let $p$ be a prime number; let us denote by $S_{1,\alpha}(X)$ the formal power series with bounded coefficients which corresponds to the measure $\mu_{1,\alpha}$. One then has $S_{1,\alpha}(0) = \langle \mu_{1,\alpha}, Q_0 \rangle = \frac{1-\alpha}{2\alpha}$.

The measure $\mu_{1,\alpha}$ is invertible in $M(\mathbb{Z}_p, \mathbb{K})$ (for the convolution product) if and only if $S_{1,\alpha}$ is invertible in the Banach algebra $K\langle X \rangle$ (for the Cauchy product). According to Theorem 3.3, the norm of measure $\mu_{1,\alpha}$ is equal to 1. Hence, $S_{1,\alpha}$ is invertible in $K\langle X \rangle$ if and only if:

$$1 = \|S_{1,\alpha}\| = |S_{1,\alpha}(0)| = \left|\frac{1-\alpha}{2\alpha}\right| = \left|\frac{1-\alpha}{2}\right|.$$

Thus $\mu_{1,\alpha}$ is invertible if and only if $\alpha - 1 \not\equiv 0 \pmod{p}$ for $p \ne 2$ (respectively $\alpha \not\equiv 1 \pmod{4}$ for $p=2$).

Since

$$(1+X)^{\alpha^{-1}} = \sum_{j \ge 0} \binom{\alpha^{-1}}{j} X^j,$$

one obtains

$$S_{1,\alpha}(X) = U_{\alpha}(X)[1 + XU_{\alpha}(X)]^{-1},$$

where

$$U_{\alpha}(X) = \alpha \sum_{j \ge 0} \binom{\alpha^{-1}}{j+2} X^j.$$

Moreover, if $\alpha$ is a $p$-adic unit such that $\alpha - 1 \ne 0 \pmod{p}$ for $p \ne 2$ and $\alpha \ne 1 \pmod{4}$ for $p=2$, one has $1 = \left|\frac{1-\alpha}{2\alpha}\right| = |U_{\alpha}(0)| \le \|U_{\alpha}\| \le 1$. One has $\|U_{\alpha}\| = |U_{\alpha}(0)| \ne 0$; hence $U_{\alpha}$ is invertible. It is readily seen that $1+XU_{\alpha}$ is invertible; one deduces that $\mu_{1,\alpha} = U_{\alpha}(\omega)[1+\omega U_{\alpha}(\omega)]^{-1}$. Thus, the convolution inverse $\nu_{\alpha}$ of the measure $\mu_{1,\alpha}$ is then given by $\nu_{\alpha} = U_{\alpha}(\omega)^{-1}[1+\omega U_{\alpha}(\omega)] = \omega + U_{\alpha}(\omega)^{-1}$. Setting

$$c_j(\alpha) = \binom{\alpha^{-1}}{j+2}$$
for $j \ge 0$, $b_0(\alpha) = 0$ and $b_j(\alpha) = c_0(\alpha)^{-1}c_j(\alpha)$ for $j \ge 1$, one has

$$
\begin{align*}
U_{\alpha}(\omega)^{-1} &= \alpha^{-1}c_0(\alpha)^{-1}\left[1 + \sum_{n \ge 1} b_n(\alpha)\omega^n\right]^{-1} \\
&= \alpha^{-1}c_0(\alpha)^{-1}\sum_{j \ge 0} (-1)^j \left[\sum_{n \ge 1} b_n(\alpha)\omega^n\right]^j \\
&= \frac{2\alpha}{1-\alpha} + \frac{2\alpha}{1-\alpha}\sum_{j \ge 1} \sum_{n \ge 1} (-1)^j b_n(j, \alpha)\omega^n,
\end{align*}
$$

with $b_n(j, \alpha) = \sum_{i_1+\cdots+i_j=n} b_{i_1}(\alpha) \cdots b_{i_j}(\alpha)$.

Since $b_0(\alpha) = 0$,

$$
b_n(j, \alpha) = \sum_{i_1 + \cdots + i_j = n} b_{i_1}(\alpha) \cdots b_{i_j}(\alpha) = 0
$$

for $j \ge n + 1$, one has

$$
b_n(j, \alpha) = \sum_{\substack{i_1, \dots, i_j \ge 1 \\ i_1 + \dots + i_j = n}} b_{i_1}(\alpha) \dots b_{i_j}(\alpha), \text{ for } j \le n.
$$

More precisely, one has

$$
b_n(j, \alpha) = \left(\frac{2\alpha^2}{1-\alpha}\right)^n \sum_{\substack{i_1, \dots, i_j \in \{1, \dots, n\} \\ i_1 + \dots + i_j = n}} \binom{\alpha^{-1}}{i_1+2} \cdots \binom{\alpha^{-1}}{i_j+2}, \text{ for } j \le n.
$$

It follows that

$$
U_{\alpha}(\omega)^{-1} = \frac{2\alpha}{1-\alpha} + \frac{2\alpha}{1-\alpha} \sum_{n \ge 1} \sum_{j=1}^{n} (-1)^j b_n(j, \alpha) \omega^n
$$

and

$$
\nu_{\alpha} = \frac{2\alpha}{1-\alpha}\delta_0 + \left[1 - \frac{2\alpha}{1-\alpha}b_1(1,\alpha)\right]\omega + \frac{2\alpha}{1-\alpha}\sum_{n \ge 2} \sum_{j=1}^{n} (-1)^j b_n(j,\alpha)\omega^n.
$$

As $b_1(1, \alpha) = b_1(\alpha) = \binom{\alpha^{-1}}{2}^{-1} \binom{\alpha^{-1}}{3} = \frac{1-2\alpha}{3\alpha}$, one obtains

$$
\begin{align*}
\nu_{\alpha} &= \frac{2\alpha}{1-\alpha}\delta_0 + \frac{1+\alpha}{3(1-\alpha)}\omega + \frac{2\alpha}{1-\alpha}\sum_{n \ge 2} \left[ \sum_{j=1}^{n} (-1)^j b_n(j, \alpha) \right] \omega^n \\
&= \sum_{n \ge 0} d_n(\alpha) \omega^n
\end{align*}
$$
where $d_0(\alpha) = \frac{2\alpha}{1-\alpha}$, $d_1(\alpha) = \frac{1+\alpha}{3(1-\alpha)}$ and for $n \ge 2$:

$$d_n(\alpha) = \frac{2\alpha}{1-\alpha} \sum_{j=1}^{n} (-1)^j b_n(j, \alpha) \\ = \alpha^n \left(\frac{2\alpha}{1-\alpha}\right)^{n+1} \sum_{\substack{1 \le j \le n \\ i_1 + \cdots + i_j = n \\ i_1, \dots, i_j \in \{1, \dots, n\}}} (-1)^j \binom{\alpha^{-1}}{i_1+2} \cdots \binom{\alpha^{-1}}{i_j+2}$$

□

## References

[1] B. DIARRA – “Base de Mahler et autres”, in *Séminaire d’Analyse, 1994-1995 (Aubière)*, Sémin. Anal. Univ. Blaise Pascal (Clermont II), vol. 10, Univ. Blaise Pascal (Clermont II), Clermont-Ferrand, 1997, p. Exp. No. 16, 18.

[2] ________, “Cours d’analyse p-adique”, Tech. report, Université de Bamako, Faculté des Sciences et Techniques, 1999 - 2000, http://math.univ-bpclermont.fr/diarra.

[3] N. KOBLITZ – *p-adic numbers, p-adic analysis and zeta-functions*, Springer-Verlag, New York - Heidelberg - Berlin, 1977.

[4] A. F. MONNA & T. A. SPRINGER – “Intégration non-archimédienne. I”, *Nederl. Akad. Wetensch. Proc. Ser. A 66=Indag. Math. 25* (1963), p. 634–642.

[5] ________, “Intégration non-archimédienne. II”, *Nederl. Akad. Wetensch. Proc. Ser. A 66=Indag. Math. 25* (1963), p. 643–653.

[6] A. C. M. VAN ROOIJ – *Non-archimedean functional analysis*, M. Dekker, New York and Basel, 1978.

[7] W. H. SCHIKHOF – *Ultrametric calculus - an introduction to p-adic analysis*, Cambridge University Press, Cambridge, 1984.
HAMADOUN MAÏGA

Département de Mathématiques
et d'Informatique

Faculté des Sciences et Techniques

Université de Bamako

Bamako Mali

hamadounmg@yahoo.fr