Codex @codex  0

For this non-Archimedean local field, the Schwartz-Bruhat space $\mathcal{S}(F)$ consists of locally constant, compactly supported complex-valued functions. Choose a nontrivial continuous additive character $\psi :F\to {\mathbb{C}}^{\times }$ of modulus one and an additive Haar measure $dx$. We use the plus-sign convention for the Fourier transform over a local field:Compact support makes the integral absolutely convergent. The additive character and the scale of the Haar measure are part of the definition; without them there is no canonical numerical transform.

Let $q=|{\mathcal{O}}_F/{\pi }_F{\mathcal{O}}_F|$, and write $V=\mathrm{vol}({\mathcal{O}}_F)$. Define the integer $c$ by the annihilator of the valuation ring ${\mathcal{O}}_F^{\perp }={\pi }_F^c{\mathcal{O}}_F$. Equivalently, $\psi$ is trivial on ${\pi }_F^c{\mathcal{O}}_F$ and not on ${\pi }_F^{c-1}{\mathcal{O}}_F$. Such a conductor exists: continuity puts the image of some additive ball inside an arc containing no nontrivial circle subgroup, so the character is trivial on that ball; nontriviality bounds the possible ball exponents below. Then $H_n={\pi }_F^n{\mathcal{O}}_F$ has volume $V{q}^{-n}$ and character annihilator $H_n^{\perp }={\pi }_F^{c-n}{\mathcal{O}}_F$.

Translation givesThe integral is the volume if $y\in H_n^{\perp }$. Otherwise translate it by $t_0\in H_n$ with $\psi (t_{0}y)\ne 1$; Haar measure invariance multiplies the same integral by a nonidentity scalar, so it must vanish. ThusEvery Schwartz-Bruhat function is a finite linear combination of such coset indicators: compactness of its support supplies a common sufficiently small translation subgroup on which it is constant, and finitely many of its cosets cover the support. The transform of each indicator has compact support and is locally constant, since $\psi$ has open kernel. Consequently $\hat{f}\in \mathcal{S}(F)$ for every $f\in \mathcal{S}(F)$.

Apply the transform again to an indicator. The same character-orthogonality calculation, together with $(H_n^{\perp }{)}^{\perp }=H_n$, givesChoose the self-dual Haar measure, characterized here by $V=q^{c/2}$. By linearity the desired Fourier inversion isIn particular, one may rescale any nontrivial additive character to make $c=0$, and then take $V=1$. For a concrete construction, start with ${\psi }_0(x)={\psi }_{{\mathbb{Q}}_p}({\mathrm{Tr}}_{F/{\mathbb{Q}}_p}x)$, where the standard rational additive character has kernel ${\mathbb{Z}}_p$. Its annihilator is the inverse different ${\mathfrak{D}}_{F/{\mathbb{Q}}_p}^{-1}$. If that ideal is ${\pi }_F^{-d}{\mathcal{O}}_F$, the rescaled character $\psi (x)={\psi }_0({\pi }_F^{-d}x)$ has ${\mathcal{O}}_F^{\perp }={\mathcal{O}}_F$. This supplies the stated normalization and also explains how the different ideal enters local Fourier analysis.

Relative to a nontrivial additive character, the self-dual additive Haar measure makes the twice-applied Fourier transform over a local field equal to reflection. If the annihilator of the valuation ring is ${\pi }^c{\mathcal{O}}_F$ and its residue field has $q$ elements, the normalization is $\mathrm{vol}({\mathcal{O}}_F)=q^{c/2}$. Volumes of a compact open subgroup and its character annihilator then multiply to one.