# Harmonic analysis on the $p$-adic numbers

Jordan Bell
March 20, 2016

Let $p$ be prime and let $N_{p}=\{0,\ldots,p-1\}$. $\mathbb{Q}_{p}\subset\prod_{\mathbb{Z}}N_{p}$. For $x\in\mathbb{Q}_{p}$,

 $x=\lim_{m\to\infty}\sum_{k\leq m}x(k)p^{k}=\sum_{k\in\mathbb{Z}}x(k)p^{k}=\sum% _{k\geq v_{p}(x)}x(k)p^{k}$

for

 $v_{p}(x)=\inf\{k\in\mathbb{Z}:x(k)\neq 0\}.$
 $\mathbb{Z}_{p}=\{x\in\mathbb{Q}_{p}:v_{p}(x)\geq 0\}.$

For $x,y\in\mathbb{Q}_{p}$,

 $v_{p}(xy)=v_{p}(x)+v_{p}(y),\qquad v_{p}(x+y)\geq\min(v_{p}(x),v_{p}(y)),$

and $v_{p}(x)=\infty$ 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 $v_{p}(f)\geq v_{p}(g)$ 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 $p^{n}\mathbb{Z}_{p}$, $n\geq 0$. 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 $n\geq 1$, $\mathbb{Z}_{p}/p^{n}\mathbb{Z}_{p}$ is isomorphic as a ring with $\mathbb{Z}/p^{n}\mathbb{Z}$.

 $|x|_{p}=p^{-v_{p}(x)},\qquad d_{p}(x,y)=|x-y|_{p}.$

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 $\mathbb{Q}_{p}$.

 $[x]_{p}=\sum_{k\geq 0}x(k)p^{k}\in\mathbb{Z}_{p},\quad\{x\}_{p}=\sum_{k<0}x(k)% p^{k}\in[0,1)\cap\mathbb{Z}[1/p].$
 $\psi_{p}(x)=e^{2\pi i\{x\}_{p}}$

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

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

the Prüfer $p$-group, and its kernel is $\mathbb{Z}_{p}$. $\mathbb{Q}_{p}/\mathbb{Z}_{p}$ and $\mathbb{Z}[p^{\infty}]$ are isomorphic as discrete abelian groups. There is a complete algebraically closed nonarchimedean valued field $\mathbb{C}_{p}$, 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

 $\left\langle x,\xi\right\rangle=\xi(x)=\prod_{k\in\mathbb{Z}}\xi(x(k)p^{k})=% \prod_{k\in\mathbb{Z}}\xi(p^{k})^{x(k)}.$ (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

 $\ker\xi_{y}=\{x\in\mathbb{Q}_{p}:|x|_{p}\leq|y|_{p}^{-1}\}$

where $|0|_{p}^{-1}=\infty$. If $y\neq 0$ then

 $\ker\xi_{y}=\{x\in\mathbb{Q}_{p}:|x|_{p}\leq|y|_{p}^{-1}\}=p^{-v_{p}(y)}% \mathbb{Z}_{p}.$

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.11 1 Gerald B. Folland, A Course in Abstract Harmonic Analysis, p. 92, Lemma 4.9.

###### Lemma 1.

If $\xi\in\widehat{\mathbb{Q}}_{p}$ then there is some $n\in\mathbb{Z}$ such that $\left\langle x,\xi\right\rangle=1$ for $x\in p^{n}\mathbb{Z}_{p}$.

###### Proof.

Let $U=\{e^{2\pi i\theta}:|\theta|<\frac{1}{4}\}$, 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 $\xi\in\widehat{\mathbb{Q}}_{p}$, $\xi\neq 1$. By (1) there is then some $k$ such that $\xi(p^{k})\neq 1$. Now, $|p^{j}|_{p}=p^{-j}\to 0$ as $j\to\infty$, so $p^{j}\to 0$ in $\mathbb{Q}_{p}$ and therefore $\xi(p^{j})\to 1$ as $j\to\infty$. Let

 $j_{\xi}-1=\max\{k\in\mathbb{Z}:\left\langle p^{k},\xi\right\rangle\neq 1\}.$

Then $\left\langle p^{j_{\xi}-1},\xi,\neq\right\rangle 1$ and $\left\langle p^{j},\xi\right\rangle=1$ for $j\geq j_{\xi}$. In particular, $j_{\xi}=0$ is equivalent with $\left\langle 1,\xi\right\rangle=1$ and $\left\langle p,\xi\right\rangle\neq 1$.22 2 Gerald B. Folland, A Course in Abstract Harmonic Analysis, p. 92, Lemma 4.10.

###### Lemma 2.

Suppose that $\xi\in\widehat{\mathbb{Q}}_{p}$ with $\left\langle 1,\xi\right\rangle=1$ and $\left\langle p^{-1},\xi\right\rangle\neq 1$. Then there are $c_{j}\in N_{p}$, $j\geq 0$, with $c_{0}\neq 0$, such that

 $\left\langle p^{-k},\xi\right\rangle=\exp\left(2\pi i\sum_{j=1}^{k}c_{k-j}p^{-% j}\right),\qquad k\geq 1.$
###### Proof.

Let $\omega_{0}=\left\langle 1,\xi\right\rangle=1$ and for $k\geq 1$ let $\omega_{k}=\left\langle p^{-k},\xi\right\rangle\in S^{1}$, which satisfy

 $\omega_{k+1}^{p}=\left\langle p^{-k},\xi\right\rangle=\omega_{k}.$

Because $\omega_{1}^{p}=1$ this means that there is some $c_{0}\in N_{p}$ such that $\omega_{1}=e^{2\pi ic_{0}p^{-1}}$, and by hypothesis $\omega_{1}\neq 1$, which means $c_{0}\neq 0$. By induction, suppose for some $k\geq 1$ and $c_{0},\ldots,c_{k-1}\in N_{p}$, $c_{0}\neq 0$, 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 $z=e^{\frac{1}{p}i\theta}e^{2\pi icp^{-1}}$. Thus, the fact that $\omega_{k+1}^{p}=\omega_{k}$ means that there is some $c_{k}\in N_{p}$ such that

 $\omega_{k+1}=\exp\left(\frac{1}{p}\cdot 2\pi i\sum_{j=1}^{k}c_{k-j}p^{-j}% \right)\cdot e^{2\pi ic_{k}p^{-1}}=\exp\left(2\pi i\sum_{j=1}^{k+1}c_{k+1-j}p^% {-j}\right).$

We prove a final lemma.33 3 Gerald B. Folland, A Course in Abstract Harmonic Analysis, p. 92, Lemma 4.11.

###### Lemma 3.

Suppose that $\xi\in\widehat{\mathbb{Q}}_{p}$ with $\left\langle 1,\xi\right\rangle=1$ and $\left\langle p^{-1},\xi\right\rangle\neq 1$. Then there is some $y\in\mathbb{Q}_{p}$ with $|y|_{p}=1$ and $\xi=\xi_{y}$.

###### Proof.

By Lemma 2 there are $c_{j}\in N_{p}$, $j\geq 0$, $c_{0}\neq 0$, such that

 $\left\langle p^{-k},\xi\right\rangle=\exp\left(2\pi i\sum_{j=1}^{k}c_{k-j}p^{-% j}\right),\qquad k\geq 1.$

Define $y\in\mathbb{Q}_{p}$ by $y(j)=c_{j}$ for $j\geq 0$ and $y(j)=0$ for $j<0$. As $y(0)=c_{0}\neq 0$, $|y|_{p}=1$. For $k\geq 1$ and $-k\leq j\leq-1$ we have $(p^{-k}y)(j)=y(j+k)=c_{j+k}$, and for $j<-k$ we have $(p^{-k}y)(j)=y(j+k)=0$, so

 $\{p^{-k}y\}_{p}=\sum_{j<0}(p^{-k}y)(j)p^{j}=\sum_{-k\leq j\leq-1}(p^{-k}y)(j)p% ^{j}=\sum_{-k\leq j\leq-1}c_{j+k}p^{j},$

yielding

 $\left\langle p^{-k},\xi\right\rangle=\exp\left(2\pi i\sum_{-k\leq j\leq-1}c_{k% +j}p^{j}\right)=\exp(2\pi i\{p^{-k}y\}_{p}),$

i.e. $\left\langle p^{-k},\xi\right\rangle=\psi_{p}(p^{-k}y)=\left\langle p^{-k},\xi% _{y}\right\rangle$. But $\left\langle 1,\xi\right\rangle=1$ implies that $\left\langle p^{k},\xi\right\rangle=1$ for $k\geq 0$, and because $y(k)=0$ for $k<0$,

 $\left\langle 1,\xi_{y}\right\rangle=e^{2\pi i\{y\}_{p}}=1,$

which implies that $\left\langle p^{k},\xi\right\rangle=1$ for $k\geq 0$. Therefore $\left\langle p^{k},\xi\right\rangle=\left\langle p^{k},\xi_{y}\right\rangle$ 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.44 4 Gerald B. Folland, A Course in Abstract Harmonic Analysis, p. 92, Theorem 4.12.

###### Theorem 4.

$y\mapsto\xi_{y}$ is an isomorphism of topological groups $\mathbb{Q}_{p}\to\widehat{\mathbb{Q}}_{p}$.

###### Proof.

For $x\in\mathbb{Q}_{p}$,

 $\left\langle x,\xi_{y}\xi_{z}\right\rangle=\left\langle x,\xi_{y}\right\rangle% \left\langle x,\xi_{z}\right\rangle=\psi_{p}(yx)\psi_{p}(zx)=\psi_{p}(yx+zx)=% \left\langle x,\xi_{y+z}\right\rangle,$

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 $\left\langle x,\xi_{y}\right\rangle=1$, 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 groups55 5 Karl H. Hofmann and Sidney A. Morris, The Structure of Compact Groups, 2nd revised and augmented edition, p. 669, Appendix 1. 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 $\xi\in\widehat{\mathbb{Q}}_{p}$, $\xi\neq 1$. By Lemma 1, let

 $j-1=\max\{k\in\mathbb{Z}:\left\langle p^{k},\xi\right\rangle\neq 1\},$

for which $\left\langle p^{j-1},\xi\right\rangle\neq 1$ and $\left\langle p^{j},\xi\right\rangle=1$. Define $\eta\in\widehat{\mathbb{Q}}_{p}$ by

 $\left\langle x,\eta\right\rangle=\left\langle p^{j}x,\xi\right\rangle,$

which satisfies $\left\langle 1,\eta\right\rangle=\left\langle p^{j}x,\xi\right\rangle=1$ and $\left\langle p^{-1},\eta\right\rangle=\left\langle p^{j-1},\xi\right\rangle\neq 1$. 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

 $\left\langle x,\xi_{y}\right\rangle=e^{2\pi i\{yx\}_{p}}=e^{2\pi i\{z\cdot p^{% -j}x\}_{p}}=\left\langle p^{-j}x,\xi_{z}\right\rangle=\left\langle p^{-j}x,% \eta\right\rangle=\left\langle x,\xi\right\rangle,$

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

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

 $N(j,k)=\{\xi\in\widehat{\mathbb{Q}}_{p}:\textrm{|\left\langle x,\xi\right% \rangle-1|

It is a fact that $\{N(j,k):j\geq 1,k\geq 1\}$ is a local base at $1$ for the topology of $\widehat{\mathbb{Q}}_{p}$. Suppose $y\in\mathbb{Z}_{p}$. For $j\geq 1$, $k\geq 1$ and $|x|_{p}\leq p^{-k}$, we have $xy\in\mathbb{Z}_{p}$ and hence $\left\langle x,\xi_{y}\right\rangle=1$, 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 $m(K)<\infty$, (iii) if $E$ is a Borel set then

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

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

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

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 $c$, $m_{1}=cm_{2}$.66 6 Walter Rudin, Fourier Analysis on Groups, pp. 1–2.

$\mathbb{Q}_{p}$ 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, $m(\mathbb{Z}_{p})<\infty$, 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 $k\in\mathbb{Z}$,

 $\mu(p^{k}\mathbb{Z}_{p})=p^{-k}.$
###### 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 $x_{j}\in\mathbb{Z}_{p}$, $1\leq j\leq p^{k}$, such that $\mathbb{Z}_{p}=\bigcup_{1\leq j\leq p^{k}}(x_{j}+p^{k}\mathbb{Z}_{p})$, 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 $k<0$, then $p^{k}\mathbb{Z}_{p}$ is a ring and $\mathbb{Z}_{p}$ is an ideal in this ring. ∎

We calculate $\mu(x\cdot E)$.77 7 Anton Deitmar and Siegfried Echterhoff, Principles of Harmonic Analysis, second ed., p. 254, Lemma 13.2.1.

###### Lemma 6.

For $A$ a Borel set in $\mathbb{Q}_{p}$ and $x\in\mathbb{Q}_{p}$,

 $\mu(x\cdot A)=|x|_{p}\mu(A).$
###### 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 $x\neq 0$, 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 $\mu_{x}(\mathbb{Q}_{p})=\mu(\mathbb{Q}_{p})=\infty$, showing $\mu_{x}$ is not identically $0$, there is some $c_{x}>0$ such that $\mu_{x}=c_{x}\mu$.

Now, as $x\neq 0$, $v_{p}(x)\in\mathbb{Z}$ 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 $f\in L^{1}(\mathbb{Q}_{p})$ and $x\neq 0$,

 $\int_{\mathbb{Q}_{p}}f(x^{-1}y)d\mu(y)=|x|_{p}\int_{\mathbb{Q}_{p}}f(y)d\mu(y).$
###### 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.88 8 Anton Deitmar and Siegfried Echterhoff, Principles of Harmonic Analysis, second ed., p. 255, Proposition 13.2.2.

###### Theorem 8.

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

###### 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

 $\displaystyle\int_{\mathbb{Q}_{p}^{*}}f(y^{-1}x)\frac{1}{|x|_{p}}d\mu(x)$ $\displaystyle=\int_{\mathbb{Q}_{p}^{*}}(g_{y}\circ M_{y})(x)d\mu(x)$ $\displaystyle=\int_{\mathbb{Q}_{p}^{*}}g_{y}(x)d\mu_{y}(x)$ $\displaystyle=|y|_{p}\int_{\mathbb{Q}_{p}^{*}}g_{y}(x)d\mu(x)$ $\displaystyle=|y|_{p}\int_{\mathbb{Q}_{p}^{*}}\frac{f(x)}{|yx|_{p}}d\mu(x)$ $\displaystyle=\int_{\mathbb{Q}_{p}^{*}}f(x)\frac{1}{|x|_{p}}d\mu(x).$

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,

 $\displaystyle\nu_{0}(p^{k}\mathbb{Z}_{p}^{*})$ $\displaystyle=\int_{\mathbb{Q}_{p}^{*}}1_{p^{k}\mathbb{Z}_{p}^{*}}(x)\frac{1}{% |x|_{p}}d\mu(x)$ $\displaystyle=\int_{\mathbb{Q}_{p}^{*}}1_{\mathbb{Z}_{p}^{*}}(p^{-k}x)\frac{1}% {|p^{-k}\cdot p^{k}x|_{p}}d\mu(x)$ $\displaystyle=\int_{\mathbb{Q}_{p}^{*}}1_{\mathbb{Z}_{p}^{*}}(x)\frac{1}{|p^{k% }x|_{p}}d\mu_{p^{k}}(x)$ $\displaystyle=|p^{k}|_{p}\int_{\mathbb{Q}_{p}^{*}}1_{\mathbb{Z}_{p}^{*}}(x)% \frac{1}{|p^{k}x|_{p}}d\mu(x)$ $\displaystyle=\int_{\mathbb{Q}_{p}^{*}}1_{\mathbb{Z}_{p}^{*}}\frac{1}{|x|_{p}}% d\mu(x)$ $\displaystyle=\int_{\mathbb{Q}_{p}^{*}}1_{\mathbb{Z}_{p}^{*}}d\mu(x)$ $\displaystyle=\mu(\mathbb{Z}_{p}^{*}).$

Check that $1+p\mathbb{Z}_{p}$ is a subgroup of $\mathbb{Z}_{p}^{*}$ with index $p-1$: the sets $a+p\mathbb{Z}_{p}$, $a\in N_{p}$, $a\neq 0$, 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 $\mathbb{Z}_{p}\setminus\{0\}=\bigcup_{n\geq 0}p^{n}\mathbb{Z}_{p}^{*}$, for $\mathrm{Re}\,s>-1$,

 $\displaystyle\int_{\mathbb{Z}_{p}\setminus\{0\}}|x|_{p}^{s}d\mu(x)$ $\displaystyle=\sum_{n\geq 0}\int_{p^{n}\mathbb{Z}_{p}^{*}}|x|_{p}^{s}d\mu(x)$ $\displaystyle=\sum_{n\geq 0}p^{-ns}\mu(p^{n}\mathbb{Z}_{p}^{*})$ $\displaystyle=\sum_{n\geq 0}p^{-ns}p^{-n}\cdot\mu(\mathbb{Z}_{p}^{*})$ $\displaystyle=\sum_{n\geq 0}p^{-ns}p^{-n}\cdot\frac{p-1}{p}$ $\displaystyle=\frac{p-1}{p(1-p^{-1-s})}.$

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

 $\displaystyle\int_{\mathbb{Z}_{p}\setminus\{0\}}|x|_{p}^{s}d\nu(x)$ $\displaystyle=\sum_{n\geq 0}\int_{p^{n}\mathbb{Z}_{p}^{*}}|x|_{p}^{s}\frac{p}{% p-1}\frac{1}{|x|_{p}}d\mu(x)$ $\displaystyle=\frac{p}{p-1}\sum_{n\geq 0}\int_{p^{n}\mathbb{Z}_{p}^{*}}(p^{-n}% )^{s-1}d\mu(x)$ $\displaystyle=\frac{p}{p-1}\sum_{n\geq 0}p^{(-s+1)n}p^{-n}\cdot\frac{p-1}{p}$ $\displaystyle=\sum_{n\geq 0}p^{-ns}$ $\displaystyle=\frac{1}{1-p^{-s}}.$

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.99 9 Dorian Goldfeld and Joseph Hundley, Automorphic Representations and $L$-Functions for the General Linear Group, volume I, p. 16, Lemma 1.6.4.

###### Lemma 9.

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

 $\int_{\mathbb{Q}_{p}}1_{p^{n}\mathbb{Z}_{p}}(x)e^{-2\pi i\{x\}_{p}}d\mu(x)=% \begin{cases}p^{-n}&n\geq 0\\ 0&\textrm{otherwise}.\end{cases}$
###### Proof.

If $n\geq 0$ 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 $n<0$, 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 $x\in p^{n}\mathbb{Z}_{p}$,

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

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

###### Lemma 10.

For $n\in\mathbb{Z}$ and $y\in\mathbb{Q}_{p}$,

 $\int_{\mathbb{Q}_{p}}1_{p^{n}\mathbb{Z}_{p}}(x)e^{-2\pi i\{yx\}_{p}}d\mu(x)=% \begin{cases}p^{-n}&y\in p^{-n}\mathbb{Z}_{p}\\ 0&\textrm{otherwise}.\end{cases}$
###### 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.1010 10 Dorian Goldfeld and Joseph Hundley, Automorphic Representations and $L$-Functions for the General Linear Group, volume I, p. 16, Proposition 1.6.5.

###### Lemma 11.

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

 $\int_{\mathbb{Q}_{p}}1_{p^{n}\mathbb{Z}_{p}^{*}}(x)e^{-2\pi i\{x\}_{p}}d\mu(x)% =\begin{cases}p^{-n}(1-p^{-1})&n\geq 0\\ -1&n=-1\\ 0&n<-1.\end{cases}$
###### 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

 $\displaystyle\int_{\mathbb{Q}_{p}}1_{p^{n}\mathbb{Z}_{p}^{*}}(x)e^{-2\pi i\{x% \}_{p}}d\mu(x)$ $\displaystyle=\int_{\mathbb{Q}_{p}}1_{p^{n}\mathbb{Z}_{p}}(x)e^{-2\pi i\{x\}_{% p}}d\mu(x)$ $\displaystyle-\int_{\mathbb{Q}_{p}}1_{p^{n+1}\mathbb{Z}_{p}}(x)e^{-2\pi i\{x\}% _{p}}d\mu(x)$ $\displaystyle=I_{1}-I_{2}.$

We apply Lemma 9. If $n\geq 0$ 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 $n+1\geq 0$ so $I_{2}=p^{-n-1}=1$ hence $I=-1$. Finally if $n<-1$ 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

 $\widehat{f}(y)=(\mathscr{F}f)(y)=\int_{\mathbb{Q}_{p}}f(x)e^{-2\pi i\{yx\}_{p}% }d\mu(x).$

Let $\mathscr{S}$ be the set of locally constant functions $\mathbb{Q}_{p}\to\mathbb{C}$ with compact support. We call an element of $\mathscr{S}$ a $p$-adic Schwartz function.1111 11 cf. A. A. Kirillov and A. D. Gvishiani, Theorems and Problems in Functional Analysis, p. 210, no. 639. We prove that the Fourier transform of a $p$-adic Schwartz function is itself a $p$-adic Schwartz function.1212 12 Dorian Goldfeld and Joseph Hundley, Automorphic Representations and $L$-Functions for the General Linear Group, volume I, p. 17, Theorem 1.6.8.

###### Theorem 12.

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

###### 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,

 $\displaystyle\widehat{1}_{N}(y)$ $\displaystyle=\int_{\mathbb{Q}_{p}}1_{a+p^{n}\mathbb{Z}_{p}}(x)e^{-2\pi i\{yx% \}_{p}}d\mu(x)$ $\displaystyle=\int_{\mathbb{Q}_{p}}1_{p^{n}\mathbb{Z}_{p}}(-a+x)e^{-2\pi i\{y(% -a+x)+ay\}_{p}}d\mu(x)$ $\displaystyle=e^{-2\pi i\{ay\}_{y}}\int_{\mathbb{Q}_{p}}1_{p^{n}\mathbb{Z}_{p}% }(-a+x)e^{-2\pi i\{y(-a+x)\}_{p}}d\mu(x)$ $\displaystyle=e^{-2\pi i\{ay\}_{y}}\int_{\mathbb{Q}_{p}}1_{p^{n}\mathbb{Z}_{p}% }(x)e^{-2\pi i\{yx\}_{p}}d\mu(x)$ $\displaystyle=e^{-2\pi i\{ay\}_{y}}p^{-n}1_{p^{-n}\mathbb{Z}_{p}}(y).$