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For an additive character $\psi$ of a non-Archimedean local field $F$,For the character induced from the standard character of ${\mathbb{Q}}_p$, this annihilator is the inverse different.

For a nontrivial additive character $\psi$ on a local field, the character annihilator of an additive subgroup $H$ is $H^{\perp }=\{y:\psi (xy)=1\text{ for all }x\in H\}$. If ${\mathcal{O}}_F^{\perp }={\pi }^c{\mathcal{O}}_F$, then $({\pi }^n{\mathcal{O}}_F{)}^{\perp }={\pi }^{c-n}{\mathcal{O}}_F$. This is a character pairing construction, distinct from the annihilator of a module.

For a scalar function on a finite abelian group, its coefficient at an additive character $\chi$ is the displayed uniform expectation. On ${\mathbb{F}}_3^d$, write ${\chi }_{\xi }(x)=e^{2\pi i\xi \cdot x/3}$. The coefficient at the constant character is the mean of $f$, and the Parseval identity on a finite group becomes ${\sum }_{\chi }|\hat{f}(\chi ){|}^2=\mathbb{E}|f{|}^2$ after relabelling characters to match the transform convention.

For a compact open additive subgroup $H$ of a local field, translation and additive character orthogonality give the displayed transform of a coset indicator. With the self-dual Haar measure, transforming again gives $1_{a+H}(-x)$. Every Schwartz-Bruhat function is a finite linear combination of these indicators.

For a local field $F$, an additive character $\psi$, and a Haar measure $dx$, the Fourier transform isIt maps the Schwartz-Bruhat space to itself.

A linear phase on an abelian group is an additive character, such as $x\mapsto e(\theta x)$ on the integers or $x\mapsto {\omega }^{rx}$ on ${\mathbb{Z}}_N$.

Write $g=(a,b)$ for the affine map $z\mapsto az+b$. Conjugation by $(a,0)$ sends translation by $b$ to translation by $ab$, so it permutes transitively the nontrivial characters of $U$. Since $U$ is normal, the induced character is zero outside $U$. On a translation by $b$ it is the sum of all conjugates of $\psi$, henceFor $b\ne 0$ the sum of the nontrivial additive characters is minus one, since the sum including the trivial character is zero. Its character norm isBy the irreducibility criterion for characters, $\rho ={\mathrm{Ind}}_U^G\psi$ is irreducible of dimension $p-1$. All choices of nontrivial $\psi$ give the same representation.

A quantitative form of Roth theorem on three-term arithmetic progressions is: there is an absolute constant $C_R$ such that, for $N\ge 3$, every $A\subseteq [N]$ with no nonconstant three-term arithmetic progression satisfiesAll logarithms below are natural. We prove the classical Roth bound for three-term progressions with an unoptimized $C_R=100$. The empty set is immediate; write $\delta =|A|/N>0$.

First establish a quantitative density increment. Suppose $N\ge 4\cdot 10^6{\delta }^{-4}$. Put $M=2N+1$, embed $I=[N]$ in $G={\mathbb{Z}}_M$, and extend all interval functions by zero. Let $a=1_A$, $u=\delta 1_I$ and $f=a-u$, the balanced subset indicator. DefineThe relation in the sum is modulo $M$. For interval elements its integer discrepancy has absolute value less than $M$, so it is precisely the integer relation. Thus $\Lambda (a,a,a)=|A|=\delta N$. The full interval count is the number of endpoint pairs of the same parity:Consequently, since $N\ge 4{\delta }^{-2}$,

Orthogonality of the additive characters givesBoth identities follow by expanding the sums and using ${\sum }_r{e}^{2\pi irt/M}=M$ for $t=0$ in $G$, and zero otherwise; the latter follows from a finite geometric series. Since $M$ is odd, $r\mapsto -2r$ permutes the frequencies. Telescope the difference asThe Cauchy-Schwarz inequality and the displayed Parseval identity on a finite group bound each term by $\delta N{\max }_r|\hat{f}(r)|$. Indeed, $\Vert a{\Vert }_2^2=\delta N$ and $\Vert u{\Vert }_2^2={\delta }^{2}N$, so every product of the two remaining L2 norms is at most $\delta N$. HenceThe zero coefficient vanishes because $\sum f=0$, so choose a nonzero frequency $\theta =r/M$ giving this bound.

Let $Q=\lfloor \sqrt{N}\rfloor$. The pigeonhole principle applied to the $Q+1$ fractional parts $0,\theta ,\dots ,Q\theta$ in $Q$ intervals gives an integer $1\le q\le Q$ with $\Vert q\theta {\Vert }_{\mathbb{R}/\mathbb{Z}}\le 1/Q$. Partition each residue class modulo $q$ in $[N]$ into consecutive arithmetic progressions of exactlyelements, leaving fewer than $L$ remainder points per residue. The total remainder $R_0$ has size at most $qL\le {\delta }^{2}N/1000$. The size hypothesis gives $L\ge {\delta }^2\sqrt{N}/2000$. On a full arithmetic progression $P$, the phase $e^{-2\pi i\theta x}$ varies from its value at the first point by at most $2\pi L/Q\le 4\pi {\delta }^2/1000$.

Let $F_P={\sum }_{x\in P}f(x)$. Replacing the phase by its first value on each full arithmetic progression and estimating the remainder using $|f|\le 1$ givesAlso $|{\sum }_P{F}_P|=|{\sum }_{R_0}f|\le {\delta }^{2}N/1000$. Therefore the sum of the positive $F_P$ isSince the total length of these arithmetic progressions is at most $N$, at least one has $F_P/|P|\ge {\delta }^2/64$. Thus the step just proved isAn affine map taking this arithmetic progression to $[|P|]$ preserves the absence of nonconstant three-term arithmetic progressions.

Now iterate. Write $N_j,{\delta }_j$ for successive lengths and subset densities, with $N_0=N$ and ${\delta }_0=\delta$. Since every subset density is at most one,Consequently $T=\lceil 65/\delta \rceil$ successful steps are impossible. On the other hand, subset densities never decrease, so with $B=\log (2000/{\delta }^2)$,Suppose $\delta \log \log N>100$. Since $T\le 66/\delta$,Let $D=\log (4\cdot 10^6{\delta }^{-4})$. We have $2B+D<31+8\log (1/\delta )\le 39/\delta <e^{54/\delta }$. Therefore every length through step $T$ satisfies $\log N_j>D$, and hence $N_j\ge 4\cdot 10^6{\delta }^{-4}\ge 4\cdot 10^6{\delta }_j^{-4}$. The density increment lemma remains applicable for all $T$ steps, contradicting the reciprocal-density bound.

Thus $\delta \log \log N\le 100$, proving the boxed theorem. The proof includes the Fourier analysis on a finite abelian group detection, conversion to an integer arithmetic progression, and the length bookkeeping that produces the double logarithm.

Count a three-term arithmetic progression as an ordered triple $(x,x+d,x+2d)$, allowing $d=0$. Inserting Fourier inversion on a finite group into its normalized count and using additive character orthogonality in $x,d$ givesIndeed the two frequency constraints are $r+s+t=0$ and $s+2t=0$, so the surviving frequencies are $(r,s,t)=(t,-2t,t)$. The term at zero is ${\alpha }^3$.

For odd $N$, let $f=a-\alpha$ be the balanced indicator function of a finite subset, and put $M={\max }_{s\ne 0}|\hat{f}(s)|$. Its zero coefficient is zero, and its other coefficients equal those of $a$. Since $-2r\ne 0$ when $r\ne 0$, the Parseval identity on a finite group givesThe assumed deficit $T<{\alpha }^3/2$ therefore forcesThus $c=1/2$ and $C=2$ suffice. This is the large Fourier coefficient from a deficit of three-term progressions estimate; no generalized von Neumann theorem is needed. For the remaining prime $N=2$, the count is $T={\mathbb{E}}_{x,d}a(x)a(x+d)={\alpha }^2$, so the deficit hypothesis is impossible. The zero common difference was included throughout.

Assume $A$ is nonempty, so $0<\alpha \le 1$, and put $a=1_A$. By the Parseval identity on a finite group, ${\sum }_r|\hat{a}(r){|}^2=\alpha$. Select the nonzero frequenciesThen $|R|{\alpha }^3/4\le \alpha$, giving $|R|\le 4{\alpha }^{-2}$. Outside $R\cup \{0\}$, the fourth Fourier moment has the boundThe fourfold normalized convolution on a finite group $H=a\ast a\ast 1_{-A}\ast 1_{-A}$ has transform $|\hat{a}(r){|}^4$, so Fourier inversion on a finite group gives $H(x)={\sum }_r|\hat{a}(r){|}^4{e}_N(rx)$. It is real and nonnegative, and $H(x)>0$ implies $x\in 2A-2A$ because the convolution counts representations, with a positive normalization.

Use the Bohr set in phase-distance conventionOn this set all retained additive characters have positive real part. The zero-frequency contribution is $\hat{a}(0{)}^4={\alpha }^4$, and the total absolute contribution of the discarded frequencies is at most ${\alpha }^4/4$. HenceThereforeThis is an explicit Bogolyubov lemma. In the chord-distance convention for a Bohr set, the phase-radius $1/10$ corresponds to chord radius $2\sin (\pi /10)>1/10$; alternatively the same positivity proof works directly at chord radius $1/10$. Thus the requested lower bound on width holds in either convention. If $R$ is empty, the Bohr set is the whole group. Nonemptiness of $A$ is necessary for the density statement.

Put $e_N(t)=e^{2\pi it/N}$ and use normalized averages ${\mathbb{E}}_x=N^{-1}{\sum }_{x\in \mathbb{Z}/N\mathbb{Z}}$. Our convention for Fourier coefficients on a finite abelian group isThe underlying orthogonality of complex exponentials is ${\mathbb{E}}_x{e}_N(rx)=1$ for $r=0$ and zero otherwise. For nonzero $r$ this is the finite geometric series with ratio $e_N(r)\ne 1$; its sum vanishes because that ratio has $N$th power one. The identical orthogonality statement holds when averaging over $r$ instead of $x$.

Using this orthogonality to sum the definition over $r$ proves Fourier inversion on a finite group:Indeed $\mathbb{E}f\overline{g}={\sum }_r\overline{\hat{g}(r)}{\mathbb{E}}_{x}f(x)e_N(-rx)={\sum }_r\hat{f}(r)\overline{\hat{g}(r)}$. This proves the Parseval identity on a finite group and its squared-norm case:For normalized convolution on a finite group, write $(f\ast g)(x)={\mathbb{E}}_{y}f(y)g(x-y)$. Changing variables $x=y+z$ in its transform givesthe convolution theorem on a finite group. Linearity of the transform follows directly from its definition. Translation ${\tau }_{t}f(x)=f(x+t)$ gives $\widehat{{\tau }_{t}f}(r)=e_N(rt)\hat{f}(r)$ by changing variables. Complex conjugation gives $\hat{\overline{f}}(r)=\overline{\hat{f}(-r)}$; for $\tilde{f}(x)=\overline{f(-x)}$, the corresponding identity is $\hat{\tilde{f}}(r)=\overline{\hat{f}(r)}$. Finally $\hat{f}(0)=\mathbb{E}f$ and $|\hat{f}(r)|\le \mathbb{E}|f|$ by the triangle inequality. These identities establish the basic normalized Fourier analysis on a finite abelian group needed below, with all normalizations fixed.

For completeness, pointwise multiplication has the dual convolution formula $\widehat{fg}(r)={\sum }_s\hat{f}(s)\hat{g}(r-s)$: insert the inversion formula for $g$ into the defining average and collect its frequencies. If $c\ne 0$ modulo $N$, the change of variables $u=cx$ gives $\widehat{f(c\cdot )}(r)=\hat{f}(r{c}^{-1})$. These also follow directly from the same additive character orthogonality.

Write $e(t)=\exp (2\pi it)$. A sequence in the circle group is an equidistributed sequence if, for every interval $I\subseteq \mathbb{R}/\mathbb{Z}$,where $|I|$ is its normalized length. Equivalently, averages of every continuous function along the sequence tend to its circle integral. Trigonometric polynomials approximate continuous functions, and interval indicators can be squeezed between continuous functions with arbitrarily close integrals. The nonconstant additive characters have integral zero. These facts give the Weyl criterion:This is the link between equidistribution and cancellation in exponential sums.

Suppose every positive-shift difference sequence were equidistributed. Fix $m\ne 0$ and set $z_n=e(m\varphi (n))$. For every fixed $h\ge 1$, the Weyl criterion would giveThe omitted final $h$ terms change a normalized average by at most $h/N$.

Here is the needed Van der Corput inequality for finite scalar sequences. Extend $z_n$ by zero outside $[1,N]$ and average $H$ consecutive translates of the sum. Cauchy-Schwarz givesIndeed, apply Cauchy-Schwarz to $H{\sum }_n{z}_n={\sum }_t{\sum }_{j=0}^{H-1}{z}_{t-j}$ and expand the squared inner sum. Taking $N\to \infty$ first leaves a bound $1/H$; then let $H\to \infty$. Every nonzero Fourier average of $\varphi$ vanishes, so the Weyl criterion makes $\varphi$ equidistributed. This is the differencing obstruction to equidistribution. By contraposition, a non-equidistributed sequence has a non-equidistributed difference for some positive $h$, hence for some $h\ne 0$ as requested.

For $\varphi (n)=\sqrt{2}{n}^2$, the difference is ${\Delta }_h\varphi (n)=-2h\sqrt{2}n-h^2\sqrt{2}$. For any nonzero integer $m$, its exponential sum is a constant phase times a geometric progression with ratio $e(-2mh\sqrt{2})\ne 1$. Its normalized magnitude is at most $2/(N|1-e(-2mh\sqrt{2})|)$, which tends to zero. Thus every positive-shift difference is equidistributed, and the contraposition just established proves

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.

The Schwartz-Bruhat space $\mathcal{S}(F)$ of a non-Archimedean local field $F$ is the vector space of locally constant, compactly supported complex-valued functions on $F$. Fix a nontrivial continuous additive character $\psi :F\to {\mathbb{C}}^{\times }$ and a Haar measure $dx$. With the sign convention required in the question, the Fourier transform over a local field is

Letbe the annihilator of the valuation ring. Translation invariance givesIndeed, the integral is the volume when the character is trivial; otherwise translation by an element on which the character is nontrivial multiplies the integral by a scalar different from one, forcing it to vanish. With the usual character of conductor ${\mathcal{O}}_F$ and the normalization $\mathrm{vol}({\mathcal{O}}_F)=1$, this becomesFor the canonical character induced from ${\mathbb{Q}}_p$, the annihilator is instead the inverse different and the displayed general formula applies.

If $g(x)=f(ax+b)$ with $a\ne 0$, the substitution $u=ax+b$ and the scaling rule $dx=|a{|}^{-1}du$ give

Every locally constant compactly supported function is a finite linear combination of characteristic functions of cosets $b+a{\mathcal{O}}_F$: compactness extracts finitely many cosets on which the function is constant. The formula just proved, together with the transform of $1_{{\mathcal{O}}_F}$, shows that the transform of each such characteristic function is again locally constant and compactly supported. Therefore the Fourier transform over a local field maps $\mathcal{S}(F)$ to itself.

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.