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For the trigonometric polynomial $F(t)={\sum }_{k=1}^N{b}_{k}e(kt)$, orthogonality of complex exponentials givesIndeed, expand each square and use ${\int }_0^{1}e(jt)dt=0$ for every nonzero integer $j$, and one for $j=0$.

A linear configuration count is a weighted sum over tuples satisfying prescribed linear equations. Orthogonality of complex exponentials often rewrites such a count as an integral of Fourier transforms.

On the circle group $\mathbb{T}$, normalized Haar measure is the pushforward measure of uniform angle on $[0,2\pi )$. For a continuous function $f:\mathbb{T}\to \mathbb{C}$, defineThe factor $1/(2\pi )$ makes the integral of the constant function $1$ equal to $1$. For $g=e^{i\varphi }$, a change of angle givesThe last equality uses the $2\pi$-periodic function $u\mapsto f(e^{iu})$. Since the circle group is an abelian group, this also verifies right invariance. An arc of angular length $L$ has normalized Haar measure $L/(2\pi )$. In particular, the arc $A=\{e^{i\theta }:\pi /6\le \theta \le \pi /2\}$ and its quarter-turn $iA$ both have measure $(\pi /2-\pi /6)/(2\pi )=1/6$.

The continuous unitary characters ${\chi }_n(e^{i\theta })=e^{in\theta }$, indexed by integers, give a fully computed family of integrals:Here $n$ must be an integer, so $z^n$ is well defined and $e^{2\pi in}=1$. Every nonconstant mode cancels exactly over a full turn. For example, the integral of $f(e^{i\theta })=3+2\cos \theta -\sin (2\theta )$ is $3$, and the integral of ${\cos }^2\theta =(1+\cos (2\theta ))/2$ is $1/2$. More generally,This is the orthogonality of complex exponentials underlying the Fourier basis and Fourier series: invariant averaging separates distinct frequencies and retains the constant component.

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.

The Roth theorem on three-term arithmetic progressions says that, for each fixed $\delta >0$, a sufficiently long interval cannot have a subset of density of a finite subset at least $\delta$ with no nonconstant three-term arithmetic progression. Equivalently, the maximum size of a three-term-progression-free subset of $[N]=\{1,\dots ,N\}$ is $o(N)$. Here is a density increment proof, including its analytic ingredients.

Write $e(t)=\exp (2\pi it)$ and embed $I=[N]$ into the cyclic group $G=\mathbb{Z}/(2N+1)\mathbb{Z}$ of order $M$. For normalized Fourier analysis on a finite abelian group take $\hat{h}(r)=M^{-1}{\sum }_{x}h(x)e(-rx/M)$. The orthogonality of complex exponentials follows from a finite geometric series: its average is zero for a nonzero frequency and one for frequency zero. Expanding in these linear phases consequently proves both Fourier inversion on a finite group and the Parseval identity on a finite group. In particular,The second identity follows by summing the frequency expansion over $x,d$: the two constraints leave precisely the frequencies $(r,-2r,r)$. Since $M$ is odd, multiplication by two permutes the frequencies. Also a congruence $x+z=2y$ with $x,y,z\in I$ is an equality in the integers, since $|x+z-2y|<M$.

Let $A\subseteq I$ have subset density $\alpha =|A|/N\ge \delta$, and suppose it contains no nonconstant three-term arithmetic progression. Its balanced indicator function of a finite subset is $f=1_A-\alpha 1_I$, extended by zero outside $I$. Then $\hat{f}(0)=0$. Counting endpoints with the same parity givesIf $N\ge 4{\alpha }^{-2}$, their difference in absolute value is at least ${\alpha }^3{N}^2/(4M^2)$. Telescope the difference of the products into three terms, each containing $f$ and two factors chosen from $1_A,\alpha 1_I$. Put $s={\max }_r|\hat{f}(r)|$. The Cauchy-Schwarz inequality and the Parseval identity on a finite group bound each term by $s\alpha N/M$, since the squared normalized $L^2$ norm of either other factor is at most $\alpha N/M$. Therefore some nonzero frequency $r$ satisfiesThis is the Fourier detection of a progression-free subset of an interval step; it has just been proved rather than assumed.

We now turn this Fourier coefficient on a finite abelian group into a density increment. Set $Q=\lfloor \sqrt{N}\rfloor$. Partition the unit interval into $Q$ equal pieces and place the $Q+1$ fractional parts of $0,r/M,\dots ,Qr/M$ in them. The pigeonhole principle gives $1\le q\le Q$ with $\Vert qr/M\Vert \le 1/Q$. This proves the particular Dirichlet approximation theorem needed here. Put $\kappa ={\alpha }^2/(1000\pi )$ and $L=\lfloor \kappa Q\rfloor$. For sufficiently large $N$ depending only on $\delta$, $L\ge \kappa Q/2$ and $2L\le Q$. Each residue chain modulo $q$ in $I$ has at least $Q$ elements. Split each such chain into arithmetic progressions with between $L$ and $2L-1$ elements, absorbing its final short remainder into the previous block.

The linear phase changes by at most $4\pi L/Q\le {\alpha }^2/250$ within any block. Freezing the linear phase at the first element of each block therefore changes the preceding correlation by at most ${\alpha }^{2}N/250$, because $|f|\le 1$. The triangle inequality givesThe block sums of the balanced indicator function of a finite subset add to zero, so their positive total is half their absolute total. Some block consequently satisfiesHere $c>0$ is absolute, and shrinking it absorbs the floors. An affine parametrization of $P$ identifies $A\cap P$ with a subset of $[|P|]$ and preserves arithmetic progressions.

Repeat this Roth density-increment step. Every step raises the subset density by at least ${\delta }^2/144$, so fewer than $T=\lceil 144/{\delta }^2\rceil +1$ steps are possible. All the lengths satisfy $N_{j+1}\ge c{\delta }^2\sqrt{N_j}$ while above a threshold depending only on $\delta$. Choose the initial $N$ so large that this lower recursion stays above that threshold for $T$ steps; this is possible by working backwards through finitely many squarings. We would then force a subset density greater than one. Thus every fixed positive density eventually forces a nonconstant three-term arithmetic progression. For an infinite subset of the positive integers with positive upper asymptotic density, apply the finite conclusion along initial intervals whose subset density is bounded away from zero; this gives the usual infinite formulation as well.

For a stationary linear autoregression, causal time series means that the observation uses only current and past driving noise:This is a convergent in the sense of mean-square convergence infinite moving-average representation. The stronger usual stable-filter definition requires absolute summability; the argument below also handles the square-summable definition. Let $\Phi (z)=1-{\sum }_{k=1}^p{\varphi }_k{z}^k$ and $\Psi (z)={\sum }_{j\ge 0}{\psi }_j{z}^j$. Substituting the filter into the recurrence and comparing coefficients of the orthogonal noise gives ${\psi }_0=1$ and ${\psi }_j={\sum }_{k=1}^{\min (p,j)}{\varphi }_k{\psi }_{j-k}$. HenceThe Cauchy-Schwarz inequality ensures that $\Psi$ is an analytic function in this disk. Thus $\Phi$ has no zero strictly inside it. A boundary zero is also impossible: a zero of multiplicity $m\ge 1$ at $z_0=e^{i{\omega }_0}$ makes $|1/\Phi (r{e}^{i\omega }){|}^2$ at least a constant times $[(1-r{)}^2+(\omega -{\omega }_0{)}^2{]}^{-m}$ near that point. Its integral over $\omega$ diverges as $r\uparrow 1$. On the other hand, orthogonality of complex exponentials givesa contradiction. Therefore the causality root criterion for an autoregressive model isUnder absolute summability, the shorter boundary argument is continuity of $\Psi$ on the closed disk and the identity $\Phi \Psi =1$ there.

Write the angular dependence as $f(\varphi )={\sum }_{n\ge 1}{a}_n\sin n\varphi +{\sum }_{n\ge 0}{b}_n\cos n\varphi$, so $u^{\prime }=\sin \theta f(\varphi )e_{\varphi }$. In the spherical coordinate system, $n=e_r$ and $e_r\times e_{\varphi }=-e_{\theta }$, withThus the integral in the torque-free rotation of a spherical squirmer has componentsThe vanishing follows from north-south cancellation, including the potentially relevant first azimuthal harmonics. In the $z$ direction, orthogonality of complex exponentials givesConsequently every rotational component is determined:This assumes an integrable prescribed angular function, with the Fourier series convergent sufficiently to justify the surface integrals; a finite Fourier sum is enough for the construction below. Other harmonics can generate exterior flows even though they contribute no net rotation.

For flow-free rotation of a spherical squirmer, choose $b_0\ne 0$ and set every $a_n$ and every $b_n$ with $n\ge 1$ to zero. Then $u^{\prime }=b_0\sin \theta e_{\varphi }$, while the rigid-rotation velocity isThe total laboratory-frame boundary velocity is therefore zero everywhere. With fluid at rest at infinity, Uniqueness of Stokes flow givesA constant pressure supplies zero resultant force and torque, so this is consistent with free motion. The imposed surface actuation counter-rotates relative to the material sphere; no claim is being made that ordinary no-slip rigid rotation produces zero flow.

Write $1_A$ for the indicator function and use the normalized Fourier analysis on a finite abelian groupThere are $N^5$ sextuples satisfying the equation, since any five coordinates determine the sixth. The desired probability is consequentlyBy orthogonality of complex exponentials, the indicator of the equation isSubstitution makes all six sums independent and gives the sixth Fourier moment as a three-sum collision count:Because $1_A$ is real, $\hat{A}(-r)=\overline{\hat{A}(r)}$. Hence the probability is

For $g\in {\mathcal{H}}_N$, writeThe orthogonality of complex exponentials converts the linear configuration count intoExpand the difference between the products for $g_1$ and $g_2$ by changing one factor at a time. A typical term iswhere each $h_j$ is either $g_1$ or $g_2$. The assumed uniform norm bound controls the first factor by $\epsilon N$. The substitution $\alpha \mapsto k\alpha$ preserves an integral over the circle group, so Hölder's inequality and the three supplied $L^3$ bounds giveEach of the four terms is therefore $O(\epsilon N^3)$, and henceThis is Fourier stability of a linear configuration count.

Set $r=k-l$. Since $k$ and $l$ are even and $k\ge l+6$, one has $r\ge 6$. The holomorphic Eisenstein series in the question is absolutely convergent and decomposes asbecause every nonzero integer pair is a positive multiple of a primitive pair and the two signs contribute the factor two.

Absolute convergence, including that established in part (c) at $s=k-1$, permits Rankin–Selberg unfolding. Unfolding the Petersson inner product from the fundamental domain to the strip $0\le x<1$, $y>0$, givesThe $x$-integral uses orthogonality of complex exponentials to retain equal Fourier indices:Finally,Substitution gives the Rankin–Selberg unfolding identity for a holomorphic Eisenstein series

For $A$ in a finite abelian group and normalized Fourier transform $\widehat{1_A}$, the normalized number of sextuples satisfyingisThis follows by inserting orthogonality of complex exponentials for the displayed equation: the three variables on one side contribute $\widehat{1_A}(\gamma {)}^3$ and those on the other side contribute its complex conjugate.