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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 Cramér–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.

For a real $2\pi$-periodic square-integrable function, use the Fourier series conventionwith Fourier coefficientsThe trigonometric orthogonality relations are ${\int }_{-\pi }^{\pi }\cos nx\cos mxdx=\pi {\delta }_{nm}$ for positive $n,m$, with the same formula for sines; sine-cosine products integrate to zero. The constant has squared integral $2\pi$ and is orthogonal to every positive harmonic. Thus the squared norm of the $N$th partial sum isTaking the limit in the L2 space gives the Fourier-series Parseval identity:Square integrability is the natural hypothesis for this finite norm identity; additional smoothness can be used when pointwise convergence is required.

For this odd function, the Parseval identity from part (a) givesThe integrand is even. Expanding and integrating it givesDividing by $144$ therefore provesThis evaluates the Riemann zeta function at six by using the norm of a simple polynomial.

With the real Fourier series convention,where the Fourier coefficients areFor a square-integrable periodic function, Parseval's identity isThe Fourier coefficients require integrability, and the stated Parseval identity requires square integrability. The expansion agrees with $f$ in the L2 space sense; for piecewise smooth data it converges pointwise to the average of the two one-sided limits. Without such qualifications an arbitrary periodic function need not have a pointwise-convergent Fourier series.

Use the Fourier transform convention $\hat{f}(k)={\int }_{\mathbb{R}}f(t)e^{-ikt}dt$, with inverse factor $1/(2\pi )$. For functions in L1 space, their convolution is $h(t)={\int }_{\mathbb{R}}f(s)g(t-s)ds$. Fubini's theorem applies because the absolute double integral is $\Vert f{\Vert }_1\Vert g{\Vert }_1$. Substituting $r=t-s$ therefore proves the convolution theorem:For the specified rectangular pulse, the support condition is $|t|<1$. Its Fourier transform isThe convolution $f\ast f$ is the length of the overlap of $(-1,1)$ and $(t-1,t+1)$. Thus it is the triangular pulseThe transform's value at zero is four, given by its continuous extension. In this convention, the Parseval identity for square-integrable functions isFor the rectangular pulse the left side is two, giving the sinc-squared integralFor the triangular pulse, $\int |f\ast f{|}^{2}dt=2{\int }_0^2(2-t{)}^{2}dt=16/3$. Apply the same Parseval identity with transform $4{\sin }^{2}k/k^2$ to obtain the sinc fourth-power integralBoth integrands have removable singularities at zero and integrable tails.

With the stated Fourier transform convention, the inverse Fourier transform and the deterministic autocorrelation of a square-integrable function areAssume initially sufficient decay and smoothness, for example a Schwartz function, to justify interchanging integrations and Fourier inversion. With $s=x+y$, the cross-correlation identity isEvaluating $h$ directly and by its inverse transform at zero proves the Parseval identity:Approximation extends the identity to L2 space, as the Plancherel theorem.

For the compact cosine profile, the Fourier transform of a truncated cosine follows from $\cos x=(e^{ix}+e^{-ix})/2$:At $k=\pm 1$ the apparent singularities are removable, with value $\pi /2$. The direct squared norm is ${\int }_{-\pi /2}^{\pi /2}{\cos }^{2}xdx=\pi /2$. Hence Parseval identity and evenness givePutting $t=\pi k/2$ makes the left side ${\pi }^3/8$ times the requested integral. ThereforeAt $t=\pi /2$ the integrand's continuous extension is $1/{\pi }^2$, so no principal-value prescription is required.

The direct squared L2 norm is the length of the interval, namely $2$. The Fourier transform isFor the transformed norm, let $I={\int }_0^{\infty }{\sin }^{2}k/k^{2}dk$. Integration by parts, with vanishing endpoint term $-{\sin }^{2}k/k$, givesThe supplied oscillatory integral is understood as an improper integral. Thereforewhich agrees with the direct norm and verifies the Parseval identity with the specified normalization.

Set $g=f$. The cross-correlation at zero is $h(0)={\int }_{\mathbb{R}}|f(x){|}^{2}dx$, while part (i) gives $\hat{h}=|\hat{f}{|}^2$. Evaluating the inverse Fourier transform at zero yields the Parseval identityThis derivation is immediate for Schwartz functions, for which all integrations and Fourier inversion are justified. Approximation by such functions and completeness then extend the identity to L2 space, defining its Fourier transform by the corresponding norm limit. This is the Plancherel theorem extension; no absolute integrability of an arbitrary square-integrable function is presumed.

By Fourier inversion, the sinc function $\sin (At)/(\pi t)$ has Fourier transform equal to one for $|\omega |<A$ and zero for $|\omega |>A$. Thus the two bandwidths add to giveThe requested squared-magnitude spectrum therefore has heights four and one:Values at the jump points depend on the chosen Fourier convergence convention and do not affect any integral. The graph is shown below.

The Parseval identity with this Fourier transform normalization gives

First take a Schwartz function, so that Fourier inversion and the exchange of integrals are justified. Inserting the inverse transform of the complex conjugate givesFor $f\in L^1\cap L^2$, approximate it by Schwartz functions in both norms. The identity just proved makes their transforms a Cauchy sequence in L2 space, while convergence in L1 space makes the transforms converge uniformly to the defining transform of $f$. Thus the two transform limits agree, and passing to the $L^2$ norms proves the finite Parseval identity. This is the normalization of the Plancherel theorem appropriate to the paper.

There is a qualification to the printed hypothesis: absolute integrability alone does not guarantee a finite square integral. For instance $f(x)=x^{-3/4}$ on $(0,1)$ and zero elsewhere belongs to $L^1$ but not $L^2$. For an $L^1$ function of this kind the transform is not in $L^2$ either, by L2 density from a square-integrable Fourier transform, so both sides are infinite if the identity is interpreted in that extended sense. The finite equality requires square integrability as well.

The Parseval identity for this Fourier transform convention isApply it to $h=g$. Its squared norm is $2{\int }_0^1(1-x{)}^{2}dx=2/3$, soUnder $k=2x$, the last integrand times $dk$ becomes $2{\sin }^{4}xdx/x^4$. ConsequentlyThe integrand has continuous value $1$ at zero and decays integrably at infinity, so this is an ordinary convergent improper integral.

Insert the inverse Fourier transforms of $f$ and its complex conjugate, initially under enough decay to justify interchanging integrals:The middle step uses the stated Dirac delta distribution identity, and the last uses its sampling property. This is the Parseval identity for the chosen normalization. Approximation extends the result to square-integrable functions, including the function used below.

The squared norm of the original function isApplying the Parseval identity to part (c), using the evenness of the squared transform, givesThus the last integral equals ${\pi }^2/8$. Now set $u=\pi k/2$. Sincewe obtainThe apparent singularity at $u=\pi /2$ is removable: the continuous value of the integrand there is $1/4$. No principal-value interpretation is needed.

On the real interval, $h(x)$ is real and $H(x)=h(x{)}^2\ge 0$. Expanding and integrating the finite sum yieldsApply part (b)(iv) to the entire function $H=h^2$. Since $|H(e^{i\theta })|=|h(e^{i\theta }){|}^2$, orthogonality of complex exponentials givesConsequentlyThe quadratic form is actually nonnegative, so its absolute value can be dropped. This shifted Hilbert quadratic-form bound comes from combining the logarithmic contour estimate with a finite Parseval identity.

With $d=\Delta x$, define the semidiscrete Fourier transform byFor finite-support sequences, expand the squared modulus and use orthogonality:ConsequentlyFor a general square-summable sequence, truncate it to $|j|\le N$. The identity makes the transformed truncations Cauchy in $L^2$, and completeness defines the transform as their $L^2$ limit. Passing to the limit proves the Parseval identityThe infinite series need not converge absolutely pointwise; the $L^2$ interpretation is sufficient for the stability analysis.