1
$\begingroup$

The standard Dirac delta is a generalised function (or measure, or distribution...) $\delta(x)$ which can be seen as a weak limit functions $\delta_n(x)$ spiked at the the origin, in the sense that:

$\lim_{n \rightarrow \infty} \int_{\mathbb{R}} f(x)\delta_n(x-a)dx = f(a)$

I am aware that one can define a $p$-adic Dirac measure satisfying the sifting property

$\int_{\mathbb{Z}_p} f(x) d\delta_p = f(0)$

for functions $f:\mathbb{Z}_p \rightarrow \mathbb{R}$ (or $\mathbb{C}$).

QUESTION: Can the $p$-adic Dirac measure also be seen as a weak limit of $p$-adic functions as in the real case? That is, do there exist functions $\delta_n(x):\mathbb{Z}_p \rightarrow \mathbb{R}$ such that

$\lim_{n \rightarrow \infty}\int_{\mathbb{Z}_p} f(x)\delta_n(x-a) d\mu = f(a)$

for any suitably well behaved function $f(x)$, where $d\mu$ is the Haar measure? Or if that is not the appropriate analogue, what would be?

Apologies if this is standard, but I couldn't find a reference for this in the literature after some searching. Thanks.

$\endgroup$
1
  • 2
    $\begingroup$ Set $B_{1/n}=\{x;\;\;|x|_p\leq 1/n\}$ set $v_n=\mu(B_{1/n})$, $\mu=$ Haar mesure, $\delta_n(x)=\frac{1}{v_n}I_{B_{1/n}}(x)$, where $I_A$ is the indicator function of a set $A$. $\endgroup$ Commented Apr 25, 2016 at 15:05

1 Answer 1

6
$\begingroup$

The answer is yes. Moreover:

1) For every (first countable) locally compact group the $\delta$ measure at the identity is a weak limit of (a sequence of) continuous functions. You can find this in any abstarct harmonic analysis text, eg Folland, "A course in abstract harmonic analysis" Prop 2.42. Look up: approximate identity.

2) Explicitly for $\mathbb{Z}_p$ you can take the functions $f_n=p^n\chi_{p^n\mathbb{Z}_p}$.

3) This specific construction could be generalized to any totally disconnected locally compact group: each such group has a basis of identity nbd consisting of compact open subgroups - take the normalized charcteristic functions of these. For the general case look up:van Danzig Theorem. For compact groups look up: profinite topology.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .