Harmonic analysis on the $p$-adic numbers

1 p-adic numbers

Let $p$ be prime and let $N_p=\{0,\mathrm{\dots },p-1\}$. ${\mathbb{Q}}_p\subset {\prod }_{\mathbb{Z}}{N}_p$. For $x\in {\mathbb{Q}}_p$,

$$x=\underset{m\to \mathrm{\infty }}{lim}\sum _{k\le m}x(k)p^k=\sum _{k\in \mathbb{Z}}x(k)p^k=\sum _{k\ge v_p(x)}x(k)p^k$$

for

$$v_p(x)=inf\{k\in \mathbb{Z}:x(k)\ne 0\}.$$

$${\mathbb{Z}}_p=\{x\in {\mathbb{Q}}_p:v_p(x)\ge 0\}.$$

For $x,y\in {\mathbb{Q}}_p$,

$$v_p(xy)=v_p(x)+v_p(y),v_p(x+y)\ge \min (v_p(x),v_p(y)),$$

and $v_p(x)=\mathrm{\infty }$ if and only if $x=0$. The $p$-integers ${\mathbb{Z}}_p$ with the valuation $v_p$ are a Euclidean domain: for $f,g\in {\mathbb{Z}}_p$ with $v_p(f)\ge v_p(g)$ we have $f\cdot g^{-1}\in {\mathbb{Z}}_p$. ${\mathbb{Z}}_p^{*}$ is the set of those $x\in {\mathbb{Z}}_p$ for which there is some $y\in {\mathbb{Z}}_p$ satisfying $xy=1$.

$${\mathbb{Z}}_p^{*}=\{x\in {\mathbb{Q}}_p:v_p(x)=0\}.$$

The ideals of the ring ${\mathbb{Z}}_p$ are $\{0\}$ and $p^n{\mathbb{Z}}_p$, $n\ge 0$. From this it follows that ${\mathbb{Z}}_p$ is a discrete valuation ring, a principal ideal domain with exactly one maximal ideal, namely $p{\mathbb{Z}}_p$; ${\mathbb{Z}}_p$ is the valuation ring of ${\mathbb{Q}}_p$ with the valuation $v_p$. For $n\ge 1$, ${\mathbb{Z}}_p/p^n{\mathbb{Z}}_p$ is isomorphic as a ring with $\mathbb{Z}/p^n\mathbb{Z}$.

$${|x|}_p=p^{-v_p(x)},d_p(x,y)={|x-y|}_p.$$

With the topology induced by the metric $d_p$, ${\mathbb{Q}}_p$ is a locally compact abelian group, and $({\mathbb{Q}}_p,d_p)$ is a complete metric space. $({\mathbb{Q}}_p,|\cdot {|}_p)$ is a complete nonarchimedean valued field. For $x\in {\mathbb{Q}}_p$,

$$\{x+p^n{\mathbb{Z}}_p:n\in \mathbb{Z}\}$$

is a local base at $x$ for the topology of ${\mathbb{Q}}_p$.

$${\psi }_p(x)=e^{2\pi i{\{x\}}_p}$$

is a continuous group homomorphism ${\mathbb{Q}}_p\to S^1$. Its image is the discrete abelian group

$$\mathbb{Z}[p^{\mathrm{\infty }}]=\{e^{2\pi im{p}^{-n}}:m,n\ge 0\},$$

the Prüfer $p$-group, and its kernel is ${\mathbb{Z}}_p$. ${\mathbb{Q}}_p/{\mathbb{Z}}_p$ and $\mathbb{Z}[p^{\mathrm{\infty }}]$ are isomorphic as discrete abelian groups. There is a complete algebraically closed nonarchimedean valued field ${ℂ}_p$, unique up to unique isomorphism, that is an extension of $({\mathbb{Q}}_p,|\cdot {|}_p)$.

2 Pontryagin dual

Denote by ${\widehat{\mathbb{Q}}}_p$ the Pontryagin dual of the locally compact abelian group $({\mathbb{Q}}_p,+)$. For $\xi \in {\widehat{\mathbb{Q}}}_p$ and $x\in {\mathbb{Q}}_p$,

$$x=\sum _{k\in \mathbb{Z}}x(k)p^k$$

and

$$⟨x,\xi ⟩=\xi (x)=\prod _{k\in \mathbb{Z}}\xi (x(k)p^k)=\prod _{k\in \mathbb{Z}}\xi {(p^k)}^{x(k)}.$$ (1)

For $y\in {\mathbb{Q}}_p$, define $m_y:{\mathbb{Q}}_p\to {\mathbb{Q}}_p$ by $m_y(x)=y\cdot x$, which is a continuous group homomorphism. Then ${\xi }_y={\psi }_p\circ m_y$ is a continuous group homomorphism ${\mathbb{Q}}_p\to S^1$, namely ${\xi }_y\in {\widehat{\mathbb{Q}}}_p$. The kernel of ${\xi }_y$ is $\{x\in {\mathbb{Q}}_p:yx\in {\mathbb{Z}}_p\}$, in other words

$$\ker {\xi }_y=\{x\in {\mathbb{Q}}_p:{|x|}_p\le {|y|}_p^{-1}\}$$

where ${|0|}_p^{-1}=\mathrm{\infty }$. If $y\ne 0$ then

$$\ker {\xi }_y=\{x\in {\mathbb{Q}}_p:{|x|}_p\le {|y|}_p^{-1}\}=p^{-v_p(y)}{\mathbb{Z}}_p.$$

We shall prove that $y\mapsto {\xi }_y$ is an isomorphism of topological groups ${\mathbb{Q}}_p\to {\widehat{\mathbb{Q}}}_p$. We will use the following lemma.

Lemma 1.

If $\xi \mathrm{\in }{\widehat{\mathrm{Q}}}_p$ then there is some $n\mathrm{\in }\mathrm{Z}$ such that $\mathrm{⟨}x\mathrm{,}\xi \mathrm{⟩}\mathrm{=}\mathrm{1}$ for $x\mathrm{\in }{p}^n\mathit{}{\mathrm{Z}}_p$.

Proof.

Let , which is an open set in $S^1$. As $\xi (0)\in U$ and $\{p^n{\mathbb{Z}}_p:n\in \mathbb{Z}\}$ is a local base at $0$, there is some $n\in \mathbb{Z}$ such that $p^n{\mathbb{Z}}_p\subset {\xi }^{-1}(U)$. This means that $\xi (p^n{\mathbb{Z}}_p)\subset U$, and because $\xi :{\mathbb{Q}}_p\to S^1$ is a group homomorphism, $\xi (p^n{\mathbb{Z}}_p)$ is therefore a subgroup of $S^1$ contained in $U$. But the only subgroup of $S^1$ contained in $U$ is $\{1\}$, and therefore $\xi (p^n{\mathbb{Z}}_p)=\{1\}$. ∎

Suppose $\xi \in {\widehat{\mathbb{Q}}}_p$, $\xi \ne 1$. By (1) there is then some $k$ such that $\xi (p^k)\ne 1$. Now, ${|p^j|}_p=p^{-j}\to 0$ as $j\to \mathrm{\infty }$, so $p^j\to 0$ in ${\mathbb{Q}}_p$ and therefore $\xi (p^j)\to 1$ as $j\to \mathrm{\infty }$. Let

$$j_{\xi }-1=\max \{k\in \mathbb{Z}:⟨p^k,\xi ⟩\ne 1\}.$$

Then $⟨p^{j_{\xi }-1},\xi ,\ne ⟩1$ and $⟨p^j,\xi ⟩=1$ for $j\ge j_{\xi }$. In particular, $j_{\xi }=0$ is equivalent with $⟨1,\xi ⟩=1$ and $⟨p,\xi ⟩\ne 1$.

Lemma 2.

Suppose that $\xi \mathrm{\in }{\widehat{\mathrm{Q}}}_p$ with $\mathrm{⟨}\mathrm{1}\mathrm{,}\xi \mathrm{⟩}\mathrm{=}\mathrm{1}$ and $\mathrm{⟨}{p}^{\mathrm{-}\mathrm{1}}\mathrm{,}\xi \mathrm{⟩}\mathrm{\ne }\mathrm{1}$. Then there are $c_j\mathrm{\in }{N}_p$, $j\mathrm{\ge }\mathrm{0}$, with $c_{\mathrm{0}}\mathrm{\ne }\mathrm{0}$, such that

$$⟨p^{-k},\xi ⟩=\exp \left(2\pi i\sum _{j=1}^k{c}_{k-j}{p}^{-j}\right),k\ge 1.$$

Proof.

Let ${\omega }_0=⟨1,\xi ⟩=1$ and for $k\ge 1$ let ${\omega }_k=⟨p^{-k},\xi ⟩\in S^1$, which satisfy

$${\omega }_{k+1}^p=⟨p^{-k},\xi ⟩={\omega }_k.$$

Because ${\omega }_1^p=1$ this means that there is some $c_0\in N_p$ such that ${\omega }_1=e^{2\pi i{c}_0{p}^{-1}}$, and by hypothesis ${\omega }_1\ne 1$, which means $c_0\ne 0$. By induction, suppose for some $k\ge 1$ and $c_0,\mathrm{\dots },c_{k-1}\in N_p$, $c_0\ne 0$, such that

$${\omega }_k=\exp \left(2\pi i\sum _{j=1}^k{c}_{k-j}{p}^{-j}\right).$$

Generally, if $z^p=e^{i\theta }$ then there is some $c\in N_p$ such that $z=e^{\frac{1}{p}i\theta }{e}^{2\pi ic{p}^{-1}}$. Thus, the fact that ${\omega }_{k+1}^p={\omega }_k$ means that there is some $c_k\in N_p$ such that

$${\omega }_{k+1}=\exp \left(\frac{1}{p}\cdot 2\pi i\sum _{j=1}^k{c}_{k-j}{p}^{-j}\right)\cdot e^{2\pi i{c}_k{p}^{-1}}=\exp \left(2\pi i\sum _{j=1}^{k+1}{c}_{k+1-j}{p}^{-j}\right).$$

∎

We prove a final lemma.

Lemma 3.

Suppose that $\xi \mathrm{\in }{\widehat{\mathrm{Q}}}_p$ with $\mathrm{⟨}\mathrm{1}\mathrm{,}\xi \mathrm{⟩}\mathrm{=}\mathrm{1}$ and $\mathrm{⟨}{p}^{\mathrm{-}\mathrm{1}}\mathrm{,}\xi \mathrm{⟩}\mathrm{\ne }\mathrm{1}$. Then there is some $y\mathrm{\in }{\mathrm{Q}}_p$ with ${\mathrm{|}y\mathrm{|}}_p\mathrm{=}\mathrm{1}$ and $\xi \mathrm{=}{\xi }_y$.

Proof.

By Lemma 2 there are $c_j\in N_p$, $j\ge 0$, $c_0\ne 0$, such that

$$⟨p^{-k},\xi ⟩=\exp \left(2\pi i\sum _{j=1}^k{c}_{k-j}{p}^{-j}\right),k\ge 1.$$

Define $y\in {\mathbb{Q}}_p$ by $y(j)=c_j$ for $j\ge 0$ and $y(j)=0$ for . As $y(0)=c_0\ne 0$, ${|y|}_p=1$. For $k\ge 1$ and $-k\le j\le -1$ we have $(p^{-k}y)(j)=y(j+k)=c_{j+k}$, and for we have $(p^{-k}y)(j)=y(j+k)=0$, so

yielding

$$⟨p^{-k},\xi ⟩=\exp \left(2\pi i\sum _{-k\le j\le -1}{c}_{k+j}{p}^j\right)=\exp (2\pi i{\{p^{-k}y\}}_p),$$

i.e. $⟨p^{-k},\xi ⟩={\psi }_p(p^{-k}y)=⟨p^{-k},{\xi }_y⟩$. But $⟨1,\xi ⟩=1$ implies that $⟨p^k,\xi ⟩=1$ for $k\ge 0$, and because $y(k)=0$ for ,

$$⟨1,{\xi }_y⟩=e^{2\pi i{\{y\}}_p}=1,$$

which implies that $⟨p^k,\xi ⟩=1$ for $k\ge 0$. Therefore $⟨p^k,\xi ⟩=⟨p^k,{\xi }_y⟩$ for all $k\in \mathbb{Z}$, which implies that $\xi ={\xi }_y$. ∎

We now have worked out enough to prove that $y\mapsto {\xi }_y$ is an isomorphism.

Theorem 4.

$y\mapsto {\xi }_y$ is an isomorphism of topological groups ${\mathrm{Q}}_p\mathrm{\to }{\widehat{\mathrm{Q}}}_p$.

Proof.

For $x\in {\mathbb{Q}}_p$,

$$⟨x,{\xi }_y{\xi }_z⟩=⟨x,{\xi }_y⟩⟨x,{\xi }_z⟩={\psi }_p(yx){\psi }_p(zx)={\psi }_p(yx+zx)=⟨x,{\xi }_{y+z}⟩,$$

showing that $y\mapsto {\xi }_y$ is a group homomorphism. Suppose that ${\xi }_y=1$. Then for all $x\in {\mathbb{Q}}_p$ we have $⟨x,{\xi }_y⟩=1$, i.e. $e^{2\pi i{\{yx\}}_p}=1$, i.e. ${\{yx\}}_p=0$, i.e. $yx\in {\mathbb{Z}}_p$. This implies $y=0$, showing that $y\mapsto {\xi }_y$ is injective. It remains to show that $y\mapsto {\xi }_y$ is surjective, that it is continuous, and that it is an open map. But in fact, the open mapping theorem for locally compact groups tells us that if $f:G\to H$ is a continuous group homomorphism of locally compact groups that is surjective and $G$ is $\sigma$-compact then $f$ is open. ${\mathbb{Q}}_p$ is $\sigma$-compact: ${\mathbb{Q}}_p={\bigcup }_{n\in \mathbb{Z}}{p}^n{\mathbb{Z}}_p$. So to prove the claim it suffices to prove that $y\mapsto {\xi }_y$ is surjective and continuous.

Let $\xi \in {\widehat{\mathbb{Q}}}_p$, $\xi \ne 1$. By Lemma 1, let

$$j-1=\max \{k\in \mathbb{Z}:⟨p^k,\xi ⟩\ne 1\},$$

for which $⟨p^{j-1},\xi ⟩\ne 1$ and $⟨p^j,\xi ⟩=1$. Define $\eta \in {\widehat{\mathbb{Q}}}_p$ by

$$⟨x,\eta ⟩=⟨p^{j}x,\xi ⟩,$$

which satisfies $⟨1,\eta ⟩=⟨p^{j}x,\xi ⟩=1$ and $⟨p^{-1},\eta ⟩=⟨p^{j-1},\xi ⟩\ne 1$. Thus we can apply Lemma 3: there is some $z\in {\mathbb{Q}}_p$, ${|z|}_p=1$, such that $\eta ={\xi }_z$. Now let $y=p^{-j}z\in {\mathbb{Q}}_p$, which satisfies

$$⟨x,{\xi }_y⟩=e^{2\pi i{\{yx\}}_p}=e^{2\pi i{\{z\cdot p^{-j}x\}}_p}=⟨p^{-j}x,{\xi }_z⟩=⟨p^{-j}x,\eta ⟩=⟨x,\xi ⟩,$$

from which it follows that $\xi ={\xi }_y$. Therefore $y\mapsto {\xi }_y$ is surjective.

For $j\ge 1$ and $k\ge 1$ define

It is a fact that $\{N(j,k):j\ge 1,k\ge 1\}$ is a local base at $1$ for the topology of ${\widehat{\mathbb{Q}}}_p$. Suppose $y\in {\mathbb{Z}}_p$. For $j\ge 1$, $k\ge 1$ and ${|x|}_p\le p^{-k}$, we have $xy\in {\mathbb{Z}}_p$ and hence $⟨x,{\xi }_y⟩=1$, hence $y\in N(j,k)$. This shows that $\xi ({\mathbb{Z}}_p)\subset N(j,k)$, and therefore $y\mapsto {\xi }_y$ is continuous at $0$. ∎

3 Haar measure

For a locally compact abelian group $G$, a Haar measure on $G$ is a Borel measure $m$ on $G$ such that (i) $m(x+E)=m(E)$ for each Borel set $E$ and $x\in G$, (ii) if $K$ is a compact set then , (iii) if $E$ is a Borel set then

$$m(E)=inf\{m(U):E\subset U\text{, }U\text{ open}\},$$

and (iv) if $U$ is an open set then

$$m(E)=sup\{m(K):K\subset U\text{, }K\text{ compact}\},$$

It is a fact that for any locally compact abelian group $G$ there is a Haar measure $m$ that is not identically $0$. One proves that if $U$ is an open set then $m(U)>0$ and that if $m_1,m_2$ are Haar measures that are not identically $0$ then for some positive real $c$, $m_1=c{m}_2$.

${\mathbb{Q}}_p$ is a locally compact abelian group, so there is a Haar measure $m$ on ${\mathbb{Q}}_p$ that is not identically $0$. Because ${\mathbb{Z}}_p$ is compact, , and because ${\mathbb{Z}}_p$ is open, $m({\mathbb{Z}}_p)>0$. Then let $\mu =\frac{1}{m({\mathbb{Z}}_p)}m$, which is the unique Haar measure on ${\mathbb{Q}}_p$ satisfying

$$\mu ({\mathbb{Z}}_p)=1.$$

Lemma 5.

For $k\mathrm{\in }\mathrm{Z}$,

$$\mu (p^k{\mathbb{Z}}_p)=p^{-k}.$$

Proof.

If $k>0$, then $p^k{\mathbb{Z}}_p$ is an ideal in ${\mathbb{Z}}_p$ and ${\mathbb{Z}}_p/p^k{\mathbb{Z}}_p$ is isomorphic as a ring with $\mathbb{Z}/p^k\mathbb{Z}$. So there are $x_j\in {\mathbb{Z}}_p$, $1\le j\le p^k$, such that ${\mathbb{Z}}_p={\bigcup }_{1\le j\le p^k}(x_j+p^k{\mathbb{Z}}_p)$, and the sets $x_j+p^k{\mathbb{Z}}_p$ are pairwise disjoint. Therefore

$$1=\mu ({\mathbb{Z}}_p)=\sum _{j=1}^{p^k}\mu (x_j+p^k{\mathbb{Z}}_p)=\sum _{j=1}^{p_k}\mu (p^k{\mathbb{Z}}_p)=p^k\mu (p^k{\mathbb{Z}}_p),$$

yielding $\mu (p^k{\mathbb{Z}}_p)=p^{-k}$.

If , then $p^k{\mathbb{Z}}_p$ is a ring and ${\mathbb{Z}}_p$ is an ideal in this ring. ∎

We calculate $\mu (x\cdot E)$.

Lemma 6.

For $A$ a Borel set in ${\mathrm{Q}}_p$ and $x\mathrm{\in }{\mathrm{Q}}_p$,

$$\mu (x\cdot A)={|x|}_p\mu (A).$$

Proof.

If $x=0$ then $x\cdot A=\{0\}$ and $\mu (x\cdot A)=0$ and ${|x|}_p\mu (A)=0\cdot \mu (A)=0$. (The set ${\mathbb{Q}}_p$ is infinite and $\mu$ is translation invariant, so finite sets have measure $0$.) For $x\ne 0$, write $M_x(y)=x^{-1}\cdot y$, which is an isomorphism of locally compact groups $({\mathbb{Q}}_p,+)\to ({\mathbb{Q}}_p,+)$. Let ${\mu }_x$ be the pushforward of $\mu$ by $M_x$:

$${\mu }_x(E)=\mu (M_x^{-1}E)=\mu (\{y\in {\mathbb{Q}}_p:x^{-1}y\in E\})=\mu (x\cdot E).$$

Because $M_x$ is an isomorphism, it follows that ${\mu }_x$ is a Haar measure on ${\mathbb{Q}}_p$. And because ${\mu }_x({\mathbb{Q}}_p)=\mu ({\mathbb{Q}}_p)=\mathrm{\infty }$, showing ${\mu }_x$ is not identically $0$, there is some $c_x>0$ such that ${\mu }_x=c_x\mu$.

Now, as $x\ne 0$, $v_p(x)\in \mathbb{Z}$ and ${|x|}_p=p^{-v_p(x)}$. Then $p^{-v_p(x)}x\in {\mathbb{Z}}_p^{*}$, so there is some $y\in {\mathbb{Z}}_p^{*}$ such that $x=p^{v_p(x)}y$. As $y\in {\mathbb{Z}}_p^{*}$, $y\cdot {\mathbb{Z}}_p={\mathbb{Z}}_p$ and hence $x\cdot {\mathbb{Z}}_p=p^{v_p(x)}\cdot {\mathbb{Z}}_p$. By Lemma 5, $\mu (p^{v_p(x)}\mathbb{Z})=p^{-v_p(x)}$, so

$${\mu }_x({\mathbb{Z}}_p)=\mu (x\cdot {\mathbb{Z}}_p)=\mu (p^{v_p(x)}\mathbb{Z})=p^{-v_p(x)}$$

and therefore

$$p^{-v_p(x)}=c_x\mu ({\mathbb{Z}}_p)=c_x,$$

and ${|x|}_p=p^{-v_p(x)}$ so $c_x={|x|}_p$. Therefore ${\mu }_x={|x|}_p\mu$. ∎

Lemma 7.

For $f\mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{Q}}_p\mathrm{)}$ and $x\mathrm{\ne }\mathrm{0}$,

$${\int }_{{\mathbb{Q}}_p}f(x^{-1}y)𝑑\mu (y)={|x|}_p{\int }_{{\mathbb{Q}}_p}f(y)𝑑\mu (y).$$

Proof.

${\mu }_x$ is the pushforward of $\mu$ by $M_x(y)=x^{-1}\cdot y$, and by the change of variables formula,

$${\int }_{{\mathbb{Q}}_p}f(x^{-1}y)𝑑\mu (y)={\int }_{{\mathbb{Q}}_p}(f\circ M_x)(y)𝑑\mu (y)={\int }_{{\mathbb{Q}}_p}f(y)𝑑{\mu }_x(y)={|x|}_p{\int }_{{\mathbb{Q}}_p}f(y)𝑑\mu (y).$$

∎

The restriction of $\mu$ to the Borel $\sigma$-algebra of ${\mathbb{Q}}_p^{*}={\mathbb{Q}}_p\setminus \{0\}$ is a Borel measure on ${\mathbb{Q}}_p^{*}$. We prove that the Borel measure on ${\mathbb{Q}}_p^{*}$ whose density with respect to $\mu$ is $x\mapsto \frac{1}{{|x|}_p}$ is a Haar measure.

Theorem 8.

$\frac{1}{{|x|}_p}d\mu (x)$ is a Haar measure on the multiplicative group ${\mathrm{Q}}_p^{\mathrm{*}}$.

Proof.

For $f\in C_c({\mathbb{Q}}_p^{*})$ and $y\in {\mathbb{Q}}_p^{*}$, writing $g_y(x)=\frac{f(x)}{{|yx|}_p}$, by Lemma 7 we have

${\displaystyle {\int }_{{\mathbb{Q}}_p^{*}}}f(y^{-1}x){\displaystyle \frac{1}{{|x|}_p}}𝑑\mu (x)$ $={\displaystyle {\int }_{{\mathbb{Q}}_p^{*}}}(g_y\circ M_y)(x)𝑑\mu (x)$

$={\displaystyle {\int }_{{\mathbb{Q}}_p^{*}}}{g}_y(x)𝑑{\mu }_y(x)$

$={|y|}_p{\displaystyle {\int }_{{\mathbb{Q}}_p^{*}}}{g}_y(x)𝑑\mu (x)$

$={|y|}_p{\displaystyle {\int }_{{\mathbb{Q}}_p^{*}}}{\displaystyle \frac{f(x)}{{|yx|}_p}}𝑑\mu (x)$

$={\displaystyle {\int }_{{\mathbb{Q}}_p^{*}}}f(x){\displaystyle \frac{1}{{|x|}_p}}𝑑\mu (x).$

∎

Write $d{\nu }_0(x)=\frac{1}{{|x|}_p}d\mu (x)$. For $x\in {\mathbb{Q}}_p^{*}$, $p^{-v_p(x)}x\in {\mathbb{Z}}_p^{*}$, i.e. $x\in p^{v_p(x)}{\mathbb{Z}}_p^{*}$, and ${\mathbb{Z}}_p^{*}$ is the kernel of the group homomorphism $x\mapsto v_p(x)$, ${\mathbb{Q}}_p^{*}\to \mathbb{Z}$. It follows that the sets $p^k{\mathbb{Z}}_p^{*}$, $k\in \mathbb{Z}$, are pairwise disjoint and ${\mathbb{Q}}_p^{*}={\bigcup }_{k\in \mathbb{Z}}{p}^k{\mathbb{Z}}_p^{*}$. For $k\in \mathbb{Z}$, because $p^k{\mathbb{Z}}_p^{*}$ is a compact open set in ${\mathbb{Q}}_p$ it is the case that $1_{p^k{\mathbb{Z}}_p^{*}}\in C_c({\mathbb{Q}}_p)$ so by Lemma 7,

${\nu }_0(p^k{\mathbb{Z}}_p^{*})$ $={\displaystyle {\int }_{{\mathbb{Q}}_p^{*}}}{1}_{p^k{\mathbb{Z}}_p^{*}}(x){\displaystyle \frac{1}{{|x|}_p}}𝑑\mu (x)$

$={\displaystyle {\int }_{{\mathbb{Q}}_p^{*}}}{1}_{{\mathbb{Z}}_p^{*}}(p^{-k}x){\displaystyle \frac{1}{{|p^{-k}\cdot p^{k}x|}_p}}𝑑\mu (x)$

$={\displaystyle {\int }_{{\mathbb{Q}}_p^{*}}}{1}_{{\mathbb{Z}}_p^{*}}(x){\displaystyle \frac{1}{{|p^{k}x|}_p}}𝑑{\mu }_{p^k}(x)$

$={|p^k|}_p{\displaystyle {\int }_{{\mathbb{Q}}_p^{*}}}{1}_{{\mathbb{Z}}_p^{*}}(x){\displaystyle \frac{1}{{|p^{k}x|}_p}}𝑑\mu (x)$

$={\displaystyle {\int }_{{\mathbb{Q}}_p^{*}}}{1}_{{\mathbb{Z}}_p^{*}}{\displaystyle \frac{1}{{|x|}_p}}𝑑\mu (x)$

$={\displaystyle {\int }_{{\mathbb{Q}}_p^{*}}}{1}_{{\mathbb{Z}}_p^{*}}𝑑\mu (x)$

$=\mu ({\mathbb{Z}}_p^{*}).$

Check that $1+p{\mathbb{Z}}_p$ is a subgroup of ${\mathbb{Z}}_p^{*}$ with index $p-1$: the sets $a+p{\mathbb{Z}}_p$, $a\in N_p$, $a\ne 0$, are contained in ${\mathbb{Z}}_p^{*}$ and are pairwise disjoint. This implies

$$\mu ({\mathbb{Z}}_p^{*})=(p-1)\mu (p{\mathbb{Z}}_p)=\frac{p-1}{p}.$$

Then

$$d\nu (x)=\frac{p}{p-1}\frac{1}{{|x|}_p}d\mu (x)$$

is a Haar measure on ${\mathbb{Q}}_p^{*}$ with $\nu ({\mathbb{Z}}_p^{*})=1$.

4 Integration

As ${\mathbb{Z}}_p\setminus \{0\}={\bigcup }_{n\ge 0}{p}^n{\mathbb{Z}}_p^{*}$, for $\mathrm{Re}s>-1$,

${\displaystyle {\int }_{{\mathbb{Z}}_p\setminus \{0\}}}{|x|}_p^s𝑑\mu (x)$ $={\displaystyle \sum _{n\ge 0}}{\displaystyle {\int }_{p^n{\mathbb{Z}}_p^{*}}}{|x|}_p^s𝑑\mu (x)$

$={\displaystyle \sum _{n\ge 0}}{p}^{-ns}\mu (p^n{\mathbb{Z}}_p^{*})$

$={\displaystyle \sum _{n\ge 0}}{p}^{-ns}{p}^{-n}\cdot \mu ({\mathbb{Z}}_p^{*})$

$={\displaystyle \sum _{n\ge 0}}{p}^{-ns}{p}^{-n}\cdot {\displaystyle \frac{p-1}{p}}$

$={\displaystyle \frac{p-1}{p(1-p^{-1-s})}}.$

For $\mathrm{Re}s>0$,

${\displaystyle {\int }_{{\mathbb{Z}}_p\setminus \{0\}}}{|x|}_p^s𝑑\nu (x)$ $={\displaystyle \sum _{n\ge 0}}{\displaystyle {\int }_{p^n{\mathbb{Z}}_p^{*}}}{|x|}_p^s{\displaystyle \frac{p}{p-1}}{\displaystyle \frac{1}{{|x|}_p}}𝑑\mu (x)$

$={\displaystyle \frac{p}{p-1}}{\displaystyle \sum _{n\ge 0}}{\displaystyle {\int }_{p^n{\mathbb{Z}}_p^{*}}}{(p^{-n})}^{s-1}𝑑\mu (x)$

$={\displaystyle \frac{p}{p-1}}{\displaystyle \sum _{n\ge 0}}{p}^{(-s+1)n}{p}^{-n}\cdot {\displaystyle \frac{p-1}{p}}$

$={\displaystyle \sum _{n\ge 0}}{p}^{-ns}$

$={\displaystyle \frac{1}{1-p^{-s}}}.$

It is worth remarking that this is a factor of the Euler product for the Riemann zeta function.

We will use the following when working with the Fourier transform.

Lemma 9.

For $n\mathrm{\in }\mathrm{Z}$,

Proof.

If $n\ge 0$ and $x\in p^n{\mathbb{Z}}_p$ then ${\{x\}}_p=0$ so

$${\int }_{{\mathbb{Q}}_p}{1}_{p^n{\mathbb{Z}}_p}(x)e^{-2\pi i{\{x\}}_p}𝑑\mu (x)=\mu (p^n{\mathbb{Z}}_p)=p^{-n}.$$

If , let $y=p^n\in p^n{\mathbb{Z}}_p$, for which ${\{y\}}_p=p^n$. Define $T:{\mathbb{Q}}_p\to {\mathbb{Q}}_p$ by $T(x)=-y+x$. Then, as $\mu$ is translation invariant and as $x+y\in p^n{\mathbb{Z}}_p$ if and only if $x\in p^n{\mathbb{Z}}_p$,

${\displaystyle {\int }_{{\mathbb{Q}}_p}}{1}_{p^n{\mathbb{Z}}_p}(x)e^{-2\pi i{\{x\}}_p}𝑑\mu (x)$ $={\displaystyle {\int }_{{\mathbb{Q}}_p}}(1_{p^n{\mathbb{Z}}_p}\circ T)(y+x)e^{-2\pi i{\{T(y+x)\}}_p}𝑑\mu (x)$

$={\displaystyle {\int }_{{\mathbb{Q}}_p}}{1}_{p^n{\mathbb{Z}}_p}(y+x)e^{-2\pi i{\{y+x\}}_p}𝑑\mu (x)$

$={\displaystyle {\int }_{{\mathbb{Q}}_p}}{1}_{p^n{\mathbb{Z}}_p}(x)e^{-2\pi i{\{y+x\}}_p}𝑑\mu (x)$

$=e^{-2\pi i{\{y\}}_p}{\displaystyle {\int }_{{\mathbb{Q}}_p}}{1}_{p^n{\mathbb{Z}}_p}(x)e^{-2\pi i{\{x\}}_p}𝑑\mu (x).$

Because $e^{-2\pi i{\{y\}}_p}\ne 1$, for $I=e^{-2\pi i{\{y\}}_p}I$ we have $I=0$. ∎

Lemma 10.

For $n\mathrm{\in }\mathrm{Z}$ and $y\mathrm{\in }{\mathrm{Q}}_p$,

Proof.

If $y\in p^{-n}{\mathbb{Z}}_p$ then for any $x\in p^n{\mathbb{Z}}_p$ we have $yx\in {\mathbb{Z}}_p$ and so ${\{yx\}}_p=0$ and $I=\mu (p^n{\mathbb{Z}}_p)=p^{-n}$. ∎

Another lemma.

Lemma 11.

For $n\mathrm{\in }\mathrm{Z}$,

Proof.

${\mathbb{Z}}_p^{*}={\mathbb{Z}}_p-p{\mathbb{Z}}_p$ and $p^n{\mathbb{Z}}_p^{*}=p^n{\mathbb{Z}}_p-p^{n+1}{\mathbb{Z}}_p$ and then

${\displaystyle {\int }_{{\mathbb{Q}}_p}}{1}_{p^n{\mathbb{Z}}_p^{*}}(x)e^{-2\pi i{\{x\}}_p}𝑑\mu (x)$ $={\displaystyle {\int }_{{\mathbb{Q}}_p}}{1}_{p^n{\mathbb{Z}}_p}(x)e^{-2\pi i{\{x\}}_p}𝑑\mu (x)$

$-{\displaystyle {\int }_{{\mathbb{Q}}_p}}{1}_{p^{n+1}{\mathbb{Z}}_p}(x)e^{-2\pi i{\{x\}}_p}𝑑\mu (x)$

$=I_1-I_2.$

We apply Lemma 9. If $n\ge 0$ then $I_1=p^{-n}$ and $I_2=p^{-n-1}$ so $I=p^{-n}-p^{-n-1}=p^{-n}(1-p^{-1})$. If $n=-1$ then $I_1=0$ and $n+1\ge 0$ so $I_2=p^{-n-1}=1$ hence $I=-1$. Finally if then $I_1=0$ and $I_2=0$ so $I=0$. ∎

For $f\in L^1({\mathbb{Q}}_p)$ and $y\in {\mathbb{Q}}_p$, define $\widehat{f}\in C_0({\mathbb{Q}}_p)$ by

$$\widehat{f}(y)=(ℱf)(y)={\int }_{{\mathbb{Q}}_p}f(x)e^{-2\pi i{\{yx\}}_p}𝑑\mu (x).$$

Let $𝒮$ be the set of locally constant functions ${\mathbb{Q}}_p\to ℂ$ with compact support. We call an element of $𝒮$ a $p$-adic Schwartz function. We prove that the Fourier transform of a $p$-adic Schwartz function is itself a $p$-adic Schwartz function.

Theorem 12.

If $f\mathrm{\in }\mathrm{S}$ then $\widehat{f}\mathrm{\in }\mathrm{S}$.

Proof.

Let $n\in \mathbb{Z}$, $a\in {\mathbb{Q}}_p$, and let $N=a+p^n{\mathbb{Z}}_p$. For $y\in {\mathbb{Q}}_p$, applying Lemma 10,

${\widehat{1}}_N(y)$ $={\displaystyle {\int }_{{\mathbb{Q}}_p}}{1}_{a+p^n{\mathbb{Z}}_p}(x)e^{-2\pi i{\{yx\}}_p}𝑑\mu (x)$

$={\displaystyle {\int }_{{\mathbb{Q}}_p}}{1}_{p^n{\mathbb{Z}}_p}(-a+x)e^{-2\pi i{\{y(-a+x)+ay\}}_p}𝑑\mu (x)$

$=e^{-2\pi i{\{ay\}}_y}{\displaystyle {\int }_{{\mathbb{Q}}_p}}{1}_{p^n{\mathbb{Z}}_p}(-a+x)e^{-2\pi i{\{y(-a+x)\}}_p}𝑑\mu (x)$

$=e^{-2\pi i{\{ay\}}_y}{\displaystyle {\int }_{{\mathbb{Q}}_p}}{1}_{p^n{\mathbb{Z}}_p}(x)e^{-2\pi i{\{yx\}}_p}𝑑\mu (x)$

$=e^{-2\pi i{\{ay\}}_y}{p}^{-n}{1}_{p^{-n}{\mathbb{Z}}_p}(y).$

∎