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Choose $ϵ=1/6$ and a sufficiently large absolute constant $k$. Part (iii)'s size bound givesLetand choose a maximal dissociated set $\Lambda \subseteq \Gamma$. The entropy form of the Chang theorem givesand maximality gives $\Gamma \subseteq \mathrm{Span}(\Lambda )$.

Put $B=B(\Lambda ,1/(6|\Lambda |))$. If $y\in B$ and $\gamma \in \Gamma$, expressing $\gamma$ as a product of characters in $\Lambda$ and their inverses givesFor $f=1_{-A}\ast {\mu }_A$, Parseval identity givesAlso $\hat{\mu }(\gamma )=|\widehat{{\mu }_Y}(\gamma ){|}^{2k}$. Fourier inversion theorem therefore yieldsonce $k$ is large enough.

If $\Lambda$ is empty, interpret $B$ as $G$; the same Fourier estimate, using only the second term, is even stronger.

Part (iii) gives $\Vert f\ast \mu -f{\Vert }_{\infty }\le 1/6$. Applying this at $x$ and $x+y$ and using the last estimate givesFinally $f(0)=1$. Hence $f(y)\ge 1/3>0$ for every $y\in B$, while the support of $f$ is $A-A$. Consequently

Let $B$ be the Hilbert-Schmidt operator with kernel $\beta$, so the function-on-function linear model is $Y=BX+\epsilon$. Independence and centering give $\mathbb{E}[Y\mid X]=BX$. Letand define the functional principal component scoresThe cross-covariance operator identity $C_{YX}=B{C}_X$ gives, for every ${\lambda }_j>0$,

Since $BX$ is centered, the requested integrated variance is $\mathbb{E}\Vert BX{\Vert }^2$. Applying the Karhunen–Loève expansion to $X$ and the Parseval identity in the ${\psi }_k$ basis yieldsEquivalently, if ${\rho }_{jk}=\mathrm{Corr}({\xi }_j,{\eta }_k)$, the expression is ${\sum }_k{\gamma }_k{\sum }_{j:{\lambda }_j>0}{\rho }_{jk}^2$.

Use the Fourier transform conventionThe zero mode vanishes because the compositional order parameter has zero spatial average. Parseval identity and the Fourier transform of a derivative turn the quadratic part of the dimensionless free energy intoHere ${\sum }_{\mathbf{q}}^{+}$ is the positive-wavevector sum for a real field. Each independent complex amplitude has density proportional to $e^{-G(q)|{\varphi }_{\mathbf{q}}{|}^2}$, so its elementary Gaussian integral gives the static structure factorwhenever $G(q)>0$.

The stationary points of the Brazovskii model kernel obeyBecause $\kappa <0<\gamma$, the nonzero minimum is the nonzero-wavevector soft-mode sphereAt this wavevector,The first Gaussian field theory divergence therefore occurs at

Translational invariance makes the pointwise variance independent of $\mathbf{r}$. Applying Parseval identity to the stated Fourier convention givesIn the thermodynamic limit, the reciprocal-lattice sum becomesAt each point $\varphi (\mathbf{r})$ is a centered Gaussian random variable, so the supplied absolute third moment gives

Expanding the square and using orthogonality of Fourier modes,Finite truncations justify the calculation directly, and passage to the ${\ell }^2$ limit proves Parseval identity for every square-summable sequence.

Use normalized convolution and Fourier coefficients, and define the large spectrumBy the Parseval identity and $0\le f\le 1$,so $|\Gamma |\le {ϵ}^{-2}{\delta }^{-1}$.

The convolution theorem givesIf $y\in B(\Gamma ,ϵ)$, the part over $\Gamma$ is at mostbecause $|\hat{f}(\gamma )|\le \delta$. On the complementary spectrum, $|\hat{f}(\gamma )|<ϵ\delta$ and $|\gamma (y)-1|\le 2$, so the contribution is less than $2ϵ{\delta }^2$. The required difference is therefore less than $3ϵ{\delta }^2$.

Put $F=1_A\ast 1_A$. Then $0\le F\le 1$, $\mathbb{E}F={\alpha }^2$, and the Parseval identity givesSet $d=\lfloor {\alpha }^{2}n/(8p^2)\rfloor$. If $d<2$, the asserted integer lower bound is trivial. Otherwise take $q=d$ and $ϵ=3\alpha /(4p)$ in part b. The resulting subspace $W$ hasThus $\dim W\ge n/9>d$, and we may choose a $d$-dimensional subspace $D\le W$.

For $x\in D$, part b givesApply the supplied maximal inequality to $g(x,y)=F(y+x)-F(y)$. Since $|D{|}^{1/q}=p$, it givesHence some $y$ satisfies $F(y)>su{p}_{x\in D}|F(y+x)-F(y)|$, so $F(y+x)>0$ for every $x\in D$. Since $\mathrm{supp}F=A+A$, this proves

For real $u_1,\dots ,u_j$, the preceding series construction givesIt is centered Gaussian, and Parseval identity makes its varianceThis is the variance of ${\sum }_r{u}_r{B}_{t_r}$. The Cramer-Wold theorem proves equality of the two joint distributions.

Writing $c_i(t)=⟨1_{[0,t]},f_i{⟩}_{L^2}$ gives $X_n(t)={\sum }_{i=1}^n{\alpha }_i{c}_i(t)$, henceBy Parseval identity, the variance tends to $\Vert 1_{[0,t]}{\Vert }_2^2=t$. Thus $X_n(t)\stackrel{d}{\to }N(0,t)$.

Fourier transforming the refinement equation and changing variables givesBy Parseval identity, orthonormality of the translates is equivalent toThese are precisely the Fourier coefficients of the periodization $P(t)={\sum }_k|f(t+2\pi k){|}^2$. Therefore

It is enough to consider three alternatives $A,B,C$. Encode each voter's three pairwise preferences by $(x_i,y_i,z_i)$, where $x_i=1$ means $A>B$, $y_i=1$ means $B>C$, and $z_i=1$ means $C>A$. A valid ranking excludes $(1,1,1)$ and $(-1,-1,-1)$. Independence of irrelevant alternatives gives three Boolean functions $u,v,w$ for the social comparisons. Unanimity and transitivity force $u=v=w$: fixing arbitrary $x$, taking $y=-x$ and $z$ constantly equal to $u(x)$ shows $v(-x)=-u(x)$, and cyclic symmetry gives the claim.

Choose the voters' valid rankings independently and uniformly. A social Condorcet paradox is absent exactly whenThus transitivity for every profile gives $3\mathbb{E}[u(x)u(y)]=-1$. Conditional on $x_i$, the bit $y_i$ equals $x_i$ with probability $1/3$ and differs with probability $2/3$, so $(x,y)$ has correlation $-1/3$. ThereforeThe Fourier formula, valid for negative correlation, givesby Parseval identity. Among the numbers $(-1/3{)}^m$, the unique minimum is $-1/3$, attained at $m=1$. Equality in this weighted average therefore forces all Fourier mass onto level one. Hence $u$ is a linear Boolean function with zero constant term. Such a function can have only one nonzero coefficient: otherwise varying two coordinates would make it assume more than two values. Thus $u(x)=x_j$ or $u(x)=-x_j$ for some $j$, making voter $j$ a dictator and proving Arrow theorem.

Write $q=1-p$, $\mu =q-p$, $\sigma =2\sqrt{pq}$, and ${\varphi }_S(x)={\prod }_{j\in S}(x_j-\mu )/\sigma$. The functions ${\varphi }_S$ form the p-biased product measure orthonormal basis, so the Fourier expansion is $f={\sum }_S{\hat{f}}_p(S){\varphi }_S$. The normalized discrete derivative of a Boolean function satisfiesApplying Parseval identity and then exchanging two finite sums gives

The noise operator on the Boolean hypercube acts diagonally on the same basis: $T_{\rho }{\varphi }_S={\rho }^{|S|}{\varphi }_S$. Hence the noise stability isIts derivative isTaking the right-hand value at $\rho =0$ leaves exactly the linear Fourier weight ${\sum }_i{\hat{f}}_p(\{i\}{)}^2$, while taking the left-hand value at $\rho =1$ gives $\mathbf{I}(f)$.

For a Boolean-valued $f$, each discrete derivative of a Boolean function $D_{i}f$ takes values in $\{-1,0,1\}$ and has degree at most $k-1$. If $f$ depends on coordinate $i$, then $D_{i}f$ is nonzero, so part (iii) givesSince $f$ has degree at most $k$, the Fourier formula for total influence and Parseval identity giveIf $m$ coordinates affect $f$, then $m{2}^{-(k-1)}\le \mathbf{I}(f)\le k$, so $m\le k{2}^{k-1}$. Thus $f$ is a $k{2}^{k-1}$-junta, which is the Nisan-Szegedy junta theorem.

Let $|f|\le 1$ on ${\mathbb{Z}}_N$ and put $g(x)=f(x){\omega }^{-x^2}$. If $\Vert g{\Vert }_{U^2}^4\ge c$, then the Parseval identity givesConsequently $f$ correlates with a quadratic phase: for some $r\in {\mathbb{Z}}_N$,

In a well $-a<x<a$, the normalized state proportional to $x$ on $0<x<a$ and zero elsewhere has stationary-state coefficientsThey follow from the orthonormal basis inner products. Their squared magnitudes give energy-measurement probabilities by the Born rule; splitting even and odd indices in the Parseval identity gives convergent integer-coefficient series identities. The expansion is interpreted in the square-integrable state norm.

There is an absolute constant $c>0$ such that, for every $\delta >0$ and all sufficiently large $N$, a subset $A\subseteq [N]$ whose density of a finite subset is $\delta$ and which has no nonconstant three-term arithmetic progression has an arithmetic progression $P\subseteq [N]$ satisfying

The Fourier transform of the balanced function $f=1_A-\delta 1_{[N]}$ has a coefficient of magnitude at least $c_0{\delta }^{2}N$. Indeed, the trilinear countsatisfies $\Lambda (1_A,1_A,1_A)=|A|$, whereas $\Lambda (1_{[N]},1_{[N]},1_{[N]})\gg N^2$. Expanding $1_A=\delta 1_{[N]}+f$ and using the Parseval identity and the Cauchy-Schwarz inequality to bound every error term forces $\Vert \hat{f}{\Vert }_{\infty }\gg {\delta }^{2}N$ once $N$ is sufficiently large in terms of $\delta$.

Choose $\theta$ at which this large coefficient occurs. The Dirichlet approximation theorem gives $1\le d\le \sqrt{N}$ with $\Vert d\theta {\Vert }_{\mathbb{R}/\mathbb{Z}}\le N^{-1/2}$. Partition $[N]$ into progressions of common difference $d$ and lengths between $\eta \sqrt{N}$ and $2\eta \sqrt{N}$, where $\eta >0$ is a sufficiently small multiple of ${\delta }^2$. The linear phase $e(\theta x)$ varies by $O(\eta )$ on each cell. The large Fourier coefficient then givesSince ${\sum }_{x\in [N]}f(x)=0$, the total positive discrepancy is half the total absolute discrepancy, so at least one cell has average of $f$ at least $c{\delta }^2$. This is the required density increment.

Let $(w_m)$ be the orthonormal basis of Dirichlet eigenfunctions for a strictly positive Dirichlet realization of an elliptic operator, with eigenvalues ${\lambda }_m$. ThenFor integers $k$, these domains are the Sobolev domains of powers of an elliptic Dirichlet operator. At $k=0$ the condition is just Parseval identity and imposes no boundary condition. At $k=1,2$ it describes $H_0^1$ and $H^2\cap H_0^1$, respectively. Higher $k$ impose traces of powers of $L$, not only the trace of $u$. For $L=-d^2/d{x}^2$ on $(0,\pi )$, the smooth Dirichlet function $u=x(\pi -x)$ has sine coefficients proportional to $m^{-3}$ for odd $m$, so the weighted sum diverges at $k=3$ despite its ordinary $H^3$ regularity.

For a time-independent strictly positive Dirichlet realization of an elliptic operator and $\psi \in L^2(U)$, the eigenfunction expansionsolves $u_t+Lu=0$ with homogeneous Dirichlet boundary conditions. For $t\ge \epsilon >0$, every power ${\lambda }_m^N$ times the exponential is bounded, so the series and all its time derivatives converge in all the Sobolev domains of powers of an elliptic Dirichlet operator. Elliptic regularity and the Sobolev embedding theorem give smoothness up to the spatial boundary for positive time. Parseval identity and the dominated convergence theorem give $u(t)\to \psi$ in $L^2$ as $t\downarrow 0$.

For a uniformly bounded two-level Fourier companion matrix $C(\theta )$ with distinct roots $g_1,g_2$ of modulus at most one, its powers satisfyA positive frequency-uniform gap therefore gives a uniform bound on every power. The Parseval identity then supplies spatial L2 norm stability. When the gap closes at a repeated unit root, a Jordan block can produce growth proportional to the number of steps despite both roots having modulus one.