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The cosmological continuity equation expresses conservation of dark-matter mass: $\delta$ is the density contrast, $\mathbf{v}$ is the peculiar velocity, and a prime denotes a conformal time derivative. The cosmological Euler equation expresses momentum conservation: $\mathcal{H}=a^{\prime }/a$ is the conformal Hubble rate, $\varphi$ is the peculiar gravitational potential, $\rho$ is the density, and ${\sigma }_{ij}$ is the velocity-dispersion tensor of collisionless matter. The terms are respectively Hubble drag, convective acceleration, gravity, and velocity-dispersion stress.

For curl-free flow, introduce the peculiar-velocity divergence $\theta =\nabla \cdot \mathbf{v}$ and set ${\sigma }_{ij}=0$. A Fourier transform of the nonlinear continuity term giveswith the alpha mode-coupling kernelTaking the divergence of the Euler equation gives the quadratic velocity kernelThis is the beta mode-coupling kernel; its symmetry follows from the two velocity factors.

By the spectral theorem for real symmetric matrices, write $A=O^{t}DO$ with $O$ orthogonal and $D=\mathrm{diag}({\xi }_1,\dots ,{\xi }_n)$. Part a and the tensor-product property of the Fourier transform giveThusApplying the pullback rule from part c to the orthogonal change of variables, for which $|\det O|=1$, replaces $D$ by $A$ and $D^{-1}$ by $A^{-1}$. Since determinant and signature are invariant under orthogonal conjugation,

Using multi-index notation, the Schwartz space isIts topology is generated by the displayed seminorms. The tempered distribution ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ is its continuous dual, with weak convergence defined by convergence of every pairing with a Schwartz function.

For the conventiondifferentiation under the integral and integration by parts giveThese identities bound every Schwartz seminorm of $\hat{\phi }$ by finitely many seminorms of $\phi$, proving continuity. The Fourier inversion theorem gives the continuous inverseso the Fourier transform is a continuous isomorphism of $\mathcal{S}$. By duality,defines a continuous isomorphism of ${\mathcal{S}}^{\prime }$.

For the Cauchy problem, insert the Fourier mode $u_m=e^{im\vartheta }$. The spatial symbol isIts real part is nonpositive for every $\vartheta$, so each mode has modulus $e^{t\mathrm{Re}{\lambda }_h}\le 1$. By the discrete Fourier transform and Parseval identity, the scheme is stable in the discrete $L^2$ norm.

For a statistically homogeneous and isotropic density contrast,Using the Fourier transform conventionand translational invariance giveswithAt one point,

For a spherical top-hat volume $V=4\pi R^3/3$,The overlap of two radius-$R$ balls separated by $r\le 2R$ isTherefore

For $\xi (r)=(r/r_0{)}^{-1.8}$,ThusThis calculation assumes that the observed galaxies trace the matter field with unit, scale-independent galaxy bias, and that the fitted power law remains valid throughout the top-hat separations.

When $V$ is approximately constant, $dV/d\varphi \simeq 0$, and exponential expansion gives the constant Hubble parameter $H=H_{\inf }$. Take the spatial Fourier transformSinceeach independent inflationary scalar Fourier mode satisfies

The Riemann-Lebesgue lemma states that the Fourier transform of an $L^1$ function is continuous and tends to zero at infinity.

For a Schwartz function $f$, differentiation under the integral and integration by parts giveThe Riemann-Lebesgue lemma bounds the right side by an $L^1$ seminorm of $D^{\alpha }(x^{\beta }f)$. Every Schwartz seminorm of $\hat{f}$ is therefore bounded by finitely many Schwartz seminorms of $f$. Hence the Fourier transform maps $\mathcal{S}({\mathbb{R}}^n)$ continuously into itself.

Taking the Fourier transform of a derivative givesThus, away from $\xi =0$,This formula determines the $L^2$ solution uniquely; a distribution supported only at $\xi =0$ cannot belong to $L^2$ unless it is zero.

For $|\xi |\le 1$, the Fourier transform estimate $|\hat{f}(\xi )|\le C\Vert f{\Vert }_1$ and local integrability of $|\xi {|}^{-2}$ in three dimensions control the squared norm. For $|\xi |>1$,The Plancherel theorem therefore yields $\Vert B_j{\Vert }_{H^{s+1}}\le C(\Vert f{\Vert }_1+\Vert f{\Vert }_{H^s})$. Finally the Sobolev embedding theorem in three dimensions gives $H^{s+1}\subset C^1$ when $s+1>1+3/2$, namely when $s>3/2$.

With the stated Fourier transform convention, Fourier inversion givesFor $x>0$, close the contour in the upper half-plane. Jordan lemma removes the semicircle contribution, and the only enclosed pole is $k=ip$. Its residue isThe residue theorem therefore yields

PutIf $Y<n\le Y+X$ is prime, then $n>D=X^{1/10}$. Since $f$ is supported on $[-1,1]$, the only divisor of $n$ contributing to $W(n)$ is $d=1$, and $W(n)=f(0)=1$. Every summand $W(n{)}^2$ is nonnegative, so

LetSince $g$ is the Fourier transform of $x\mapsto e^{x}f(x)$, the Fourier inversion theorem gives the Fourier representation of a smooth Selberg weightExpanding the square, interchanging the absolutely convergent sums and integrals, and usingwe obtainThe error is $O(D^2)$ because the support of $f$ restricts both divisors to $d_j\le D$. The main factor isand its Euler product is

It remains to estimate the integral using the assumed zeta-factor bound. The transform $g$ is a Schwartz function, since $e^{x}f(x)$ is smooth and compactly supported. We may therefore truncate to $|t_1|,|t_2|\le T$, losing an arbitrarily large negative power of $T$. Uniformly in the needed truncated range, the standard estimates near the pole of the Riemann zeta function giveandAfter multiplication, one net factor $(\log D{)}^{-1}$ remains. The polynomial factors in $t_1,t_2$ are integrable against the rapidly decreasing $g(t_1)g(t_2)$, and choosing $T$ as a sufficiently large power of $\log D$ makes the tails negligible. HenceSince $D=X^{1/10}$ and $D^2=X^{1/5}\ll X/\log X$,This proves the short-interval prime upper bound from a smooth divisor weight.

The Euler-Lagrange equation isAfter a spatial Fourier transform and the change to conformal time, ${\partial }_t=a^{-1}{\partial }_{\tau }$, this becomesFor the exact de Sitter spacetime scale factor $a=-1/(H\tau )$, one has $a^{\prime \prime }/a=2a^2{H}^2$. Setting $u_{\mathbf{k}}=a{\phi }_{\mathbf{k}}$ therefore cancels both the friction term and the effective mass term, leavingThis is the special cancellation for a conformally coupled scalar field.

Under the Fourier transform, the operator $A(t)={\partial }_x^3+\varphi (t){\partial }_x$ is the Fourier multiplier operatorIts symbol is purely imaginary because $\varphi$ is real. Consequently $A(t)$ is skew-adjoint on $L^2(\mathbb{R})$ with common domain $H^3(\mathbb{R})$ and generates the strongly continuous unitary groupThe Plancherel theorem gives $\Vert e^{rA(t)}u{\Vert }_2=\Vert u{\Vert }_2$, so every $A(t)\in \mathcal{G}(1,0)$. Products of the frozen groups are also unitary; hence this is a stable family of semigroup generators with constants $1,0$.

For $u\in H^3(\mathbb{R})$,Since $\varphi \in C^1(\mathbb{R})$, the map $t\mapsto A(t)$ is continuously differentiable from $H^3$ to $L^2$. All hypotheses from part f are satisfied, so an evolution family exists on every finite interval $[0,T]$.

In this commuting Fourier multiplier operator example the solution operator can also be written explicitly:Its multiplier has absolute value one, directly confirming strong continuity, the evolution family law, preservation of $H^3$, and the required derivatives. The equation combines the dispersive Airy equation with a time-dependent linear transport equation.

The angular-frequency Fourier transform and its inverse on $\mathcal{S}({\mathbb{R}}^n)$ areIt extends to a tempered distribution by duality:

Let $K$ contain the support of $u\in {\mathcal{E}}^{\prime }({\mathbb{R}}^n)$. Choose a cutoff function that equals one near $K$. It makeswell-defined, and differentiating the parameter under the pairing givesThus the Fourier transform of a compactly supported distribution is a smooth function. Since a compactly supported distribution has finite order, some $N$ and $C$ satisfyApplying this estimate to the exponential yields $|\hat{u}(\lambda )|\le C^{\prime }⟨\lambda {⟩}^N$.

Now take $v\in {\mathcal{E}}^{\prime }(X)$, multiply by a cutoff function supported in $X$ and equal to one near $\mathrm{supp}v$, and regard the result as an element of ${\mathcal{E}}^{\prime }({\mathbb{R}}^n)$. Choose $m$ so large thatis Lebesgue integrable. The inverse Fourier transform $f={\mathcal{F}}^{-1}g$ is a bounded continuous function, and the Fourier transform of a derivative givesas a distributional identity. This is the Bessel potential proof of the structure theorem for compactly supported distributions.

The function $f$ itself need not have compact support. Choose another cutoff $\rho \in \mathcal{D}(X)$ equal to one near $\mathrm{supp}v$. Then $v=\rho (1-\Delta {)}^{m}f$. Repeatedly usingexpresses $v$ as a finite sum ${\sum }_{\beta }{\partial }^{\beta }{f}_{\beta }$, where every coefficient $f_{\beta }$ is continuous and compactly supported in $X$.

Use ${\partial }_x\mapsto iq$ under the Fourier transform and the reality conditions $\varphi (-q)=\varphi (q{)}^{\ast }$ and $p(-q)=p(q{)}^{\ast }$. With $\Psi =(\varphi ,p)$, the quadratic free energy iswhereThe opposite imaginary off-diagonal entries make $G(q)$ a Hermitian matrix. Reversing the Fourier-sign convention reverses both of those signs without changing any correlator.

For $s\in \mathbb{R}$, the Sobolev space $H^s({\mathbb{R}}^n)$ consists of tempered distributions $u$ such thatFor a multi-index $\alpha$,Sincewe obtain the Sobolev derivative estimateThus differentiation is bounded and linear between the stated spaces.

Taking the Fourier transform ofgivesThe only possible solution is thereforeMoreover,This proves existence, uniqueness, and bounded dependence. Conversely, $u\mapsto (-\Delta +1)u$ maps $H^{s+2}$ boundedly to $H^s$, sois a linear isomorphism with inverse multiplier $(1+|\xi {|}^2{)}^{-1}$. This is the massive Laplacian isomorphism on Sobolev spaces.

Finally, the assumed estimate and Plancherel theorem implyThus $(u_j)$ is bounded in $H^2({\mathbb{R}}^n)$. The functions and all their first derivatives are supported in the fixed bounded set $\overline{\Omega }$. Their $H^1$ bounds give uniform translation estimatesfor $v=u_j$ and $v=D_{\ell }{u}_j$. Tight support and the Rellich-Kondrashov compactness theorem for H01, equivalently the Fourier compactness criterion, therefore provide a common subsequence for which $u_j$ and every first derivative converge strongly in $L^2$. HenceThis is compactness from bounded support and an H2 bound.

For the stated angular-frequency Fourier transform convention, the convolution theorem givesThe Fourier inversion theorem and the assumed absolute integrability therefore imply

Since $|-i|=1$ and the principal argument is $-\pi /2$,All logarithms are $-\pi i/2+2\pi in$, $n\in \mathbb{Z}$.

Take the spatial Fourier transform of the wave equation. The transformed initial-value problem isand hence

Part (c) and the convolution theorem turn the first term intoFor the second term, choose $p$ with $p^{\prime }=g$. Part (b) says $\tilde{g}=ik\tilde{p}$, soPart (c) now makes its inverse transformCombining the two terms gives the D'Alembert formula

Let $A=2V$ and $B=W-V$. Part (c) makes $A$ and $B$ independent identically distributed exponential variables of rate $\lambda$. The change of variables $(A,B)=(rs,(1-r)s)$ has Jacobian $s$, and integrating ${\lambda }^{2}s{e}^{-\lambda s}$ over $s>0$ gives density one for $0<r<1$. Therefore the uniform ratio of independent exponential variables gives

By the Fourier inversion theorem and Fubini's theorem,Thus multiplication of Fourier transforms corresponds to the convolution $h=f\ast g$, which is the convolution theorem in the convention used here.

Take the Fourier transform in $x$. The wave equation becomes the ordinary differential equationwith initial data $\tilde{u}(k,0)=\tilde{f}(k)$ and ${\partial }_t\tilde{u}(k,0)=\tilde{g}(k)$. Therefore

Part (c) and the Translation property of the Fourier transform invert the first term toFor the second term choose an antiderivative $p$ of $g$. By part (b), $\tilde{g}=ik\tilde{p}$, and henceIts inverse transform isCombining the terms gives the D'Alembert formula