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The sources are independent odd Grassmann variables and anticommute with the Dirac field. Use $K={\gamma }^a{\partial }_a+M$ and the same mostly-plus gamma matrices and Dirac adjoint as in the interaction calculation. The inverse $K_F^{-1}$ is selected with vacuum Feynman i-epsilon prescription boundary conditions. With integral kernels and spinor contractions understood, the exponent can be completed to a square:Translations preserve the Berezin integral, so the Gaussian generating functional for a Dirac field isThe positions of the sources matter. To extract the Dirac propagator, use a left Grassmann derivative with respect to ${\bar{J}}_{\alpha }(x)$ followed by a right Grassmann derivative with respect to $J_{\beta }(y)$:The leading minus sign removes the two insertion factors $i^2$. It can also be checked by differentiating the quadratic source exponential: its ordered second derivative is $-iZ[0,0]K_F^{-1}$.

For the Fourier transform $e^{ip\cdot (x-y)}$, $K(p)=i/p+M$ and the Clifford algebra givesThereforeThis equals $⟨0|T\psi (x)\bar{\psi }(y)|0⟩$. The fermionic time ordering is explicitlywith the minus sign supplied by exchanging odd fields. The poles put positive energy forward in time and negative energy backward, which is the antiparticle contribution. Finally, $K_x{S}_F(x,y)=iI{\delta }^{(4)}(x-y)$ checks the numerator and overall sign. All formulas are distributional limits with the indicated boundary prescription.

Choose $\varphi (x)=\int d^{4}p(2\pi {)}^{-4}{e}^{ip\cdot x}\tilde{\varphi }(p)$ in Euclidean signature. The inverse quadratic kernel is the smooth-cutoff scalar propagatorThe subscript records the explicit cutoff dependence implied by the condition $f_{\Lambda }(z)={\Lambda }^{2}z$ at small $z$. For $p^2+m^2\le {\Lambda }^2$, this reduces to $1/(p^2+m^2)$, the usual Euclidean scalar propagator. For $p^2+m^2\gg {\Lambda }^2$, the kernel $f_{\Lambda }$ grows rapidly and its inverse tends to zero.

High-momentum modes are strongly suppressed. For a finite smooth regulator they are not literally identically zero; that statement would require a sharp cutoff. The field variance carried by those Fourier transform modes is correspondingly negligible.

Use the Minkowski metric, path-integral weight $e^{iS}$ and the Abelian gauge theory transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\omega$. The gauge functional $F[A]=\partial \cdot A+A^2$ varies asThus the Faddeev-Popov operator is ${\mathcal{M}}_A=\square +2A\cdot \partial$. It depends on the gauge field despite the gauge group being Abelian: the ghosts interact because this gauge condition is nonlinear.

Choose the gauge-fixing fermion $\Psi =\int d^{4}x\bar{c}(F[A]+\xi h/2)$. The gauge-fixed action $S=S_{\mathrm{Maxwell}}+s\Psi$ isHere the tensor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ is distinct from the scalar gauge functional $F[A]$. With $\xi =0$, integrating over $h$ imposes the exact printed constraint. For nonzero $\xi$, eliminating $h=-F[A]/\xi$ instead givesThis version displays the additional gauge-dependent cubic and quartic gauge-field vertices as well as the ghost interaction; the strict condition is its $\xi \to 0$ limit.

For the Fourier transform convention $c(x)=\int d^{4}p(2\pi {)}^{-4}{e}^{-ipx}c(p)$, the quadratic ghost kernel is $p^2$. With the ordering $⟨c(p)\bar{c}(q)⟩$,The term $-2\bar{c}{A}^{\mu }{\partial }_{\mu }c$ has Fourier coefficient $2i{p}_{\mu }$, where $p$ is the incoming ghost momentum. Multiplication by $i$ in the Feynman rule givesThese signs refer to the displayed action, Fourier convention and ghost ordering. Reversing the ghost/antighost convention changes corresponding signs consistently. A closed ghost loop has the additional minus sign from Grassmann variables.

The quantum harmonic oscillator has $\hat{H}=({\dot{\hat{q}}}^2+{\omega }^2{\hat{q}}^2)/2$, with $\omega >0$. Using $\hat{a}|0⟩=0$ and the canonical commutation relation givesFor $q=r{e}^{is}$, the Wronskian normalization isConsequentlyEquality requires $\dot{r}=0$ and $r^2=1/(2\omega )$. The minimum-energy normalized oscillator mode is thereforeThe constant phase $\alpha$ is irrelevant. This mode also solves the oscillator equation of motion; an arbitrary squeezed mode would have larger vacuum energy.

For inflation, introduce the Mukhanov-Sasaki variable $v=z\mathcal{R}$. Since $z$ depends only on conformal time,Integrating the cross term by parts gives the canonical bulk actionup to the boundary term $-\frac{1}{2}\int d^{3}x[(z^{\prime }/z)v^2{]}_{\mathrm{boundary}}$. The resulting Euler-Lagrange field equation and its Fourier transform areAt leading order in the slow-roll approximation, retain a small positive, nearly constant $ϵ$ in $z=a\sqrt{2ϵ}$, while approximating $a=-1/(H\tau )$ and $H$ as constant. Then $z^{\prime \prime }/z\simeq 2/{\tau }^2$, soStrict exact de Sitter spacetime would have $ϵ=0$ and would not supply the nonzero curvature kinetic coefficient assumed here; the calculation is the leading quasi-de Sitter limit, not a substitution of zero into $z$.

For $-k\tau \gg 1$, the expansion correction is negligible and each canonical mode is a quantum harmonic oscillator of conformal frequency $k$. The Bunch-Davies vacuum selects its positive-frequency, minimum-energy mode $e^{-ik\tau }/\sqrt{2k}$ in that early subhorizon regime. It does not minimize an instantaneous Hamiltonian after the effective squared frequency has become negative outside the horizon.

A basis of exact solutions to the leading mode equation is $e^{-ik\tau }(1-i/(k\tau ))$ and its complex conjugate. Write the normalized Bogoliubov transformation combination asThe Bunch-Davies vacuum boundary condition sets ${\beta }_k=0$ and ${\alpha }_k=1$ up to phase. HenceSubstitution verifies the equation, and $v_k{v}_k^{\ast \prime }-v_k^{\prime }{v}_k^{\ast }=i$ verifies the Wronskian normalization.

Dividing by $z$ gives the comoving curvature perturbation varianceThis is the dimensional slow-roll curvature power spectrum in the printed normalization, with the reduced Planck mass set to one. The corresponding dimensionless cosmological power spectrum is $k^3{P}_{\mathcal{R}}/(2{\pi }^2)=H^2/(8{\pi }^2ϵ)$, which is independent of $k$ at this order. Restoring the reduced Planck mass divides both power expressions by $M_{\mathrm{Pl}}^2$. Slowly varying background quantities are evaluated near each mode's horizon exit; their variation generates the small departure from exact scale invariance.

For a single barotropic perfect fluid in general relativity, the perturbations obey $\delta P=w\delta \rho =w\bar{\rho }\delta$. This adiabatic closure is needed: constant background $P/\rho$ alone would not eliminate an independent entropy perturbation. The absence of scalar anisotropic stress allows the common potential used in Newtonian gauge in cosmology.

Substituting the density constraint into the pressure equation gives the gravitational potential evolution of a barotropic fluidSince $a\propto {\tau }^{2/(1+3w)}$, the conformal Hubble parameter is $\mathcal{H}=2/[(1+3w)\tau ]$. The bracket cancels identically, leavingFor a Fourier transform mode, ${\nabla }^2$ becomes $-k^2$.

During radiation domination, $w=1/3$ and ${\Phi }_k^{\prime \prime }+4{\Phi }_k^{\prime }/\tau +(k^2/3){\Phi }_k=0$. Set $x=k\tau /\sqrt{3}$ and write $\Phi =u/x$. The resulting equation is $u_{xx}+2u_x/x+(1-2/x^2)u=0$, so the two Spherical Bessel functions in the hint giveAt $x\ll 1$, the two solutions approach a constant and a mode proportional to ${\tau }^{-3}$. The regular adiabatic mode, normalized to its primordial potential, isAfter entry into the sound horizon, $x\gg 1$, the potential oscillates at cosmological sound speed $1/\sqrt{3}$ with envelope $x^{-2}\propto a^{-2}$. The Hubble radius and sound horizon differ by the sound-speed factor; outside the Hubble radius the regular potential is constant, while well inside it radiation supports acoustic oscillations.

During matter domination, $w=0$ gives ${\Phi }_k^{\prime \prime }+6{\Phi }_k^{\prime }/\tau =0$ at every wavenumber. HenceThe growing density mode has a constant potential both outside and inside the Hubble radius; the other potential mode decays. Pressureless matter has zero cosmological sound speed, so horizon entry does not produce the radiation acoustic decay. These formulas cover both independent solutions, while the subsequent sketches select the regular adiabatic growing mode.

For a statistically homogeneous comoving curvature perturbation, define the connected primordial bispectrum byStatistical isotropy makes $B$ depend only on the three magnitudes, which must form a triangle. A Gaussian random field has zero connected bispectrum. Canonical attractor single-field slow-roll inflation with the usual vacuum produces only slow-roll-sized primordial non-Gaussianity. A measurable signal can arise from additional light fields and nonlinear conversion of isocurvature perturbations, or from enhanced interactions such as a small sound speed, departures from an attractor, or suitable features/excited initial states. The resulting model must still reproduce the observed nearly scale-invariant power spectrum and remain under perturbative control, with acceptable backreaction and late-time conversion to the observed perturbations. Large non-Gaussianity is therefore possible but is not automatic in a viable model.

Put $c=3f_{\mathrm{NL}}/5$ and use the consistent Fourier transform convention with ${\zeta }_G(\mathbf{x})$, rather than the repeated momentum argument printed in the integrand. The quadratic part has transformThe subtraction sets the mean to zero and removes the internal contraction of a single quadratic factor. At first order in $f_{\mathrm{NL}}$, choose one of the three external factors to be quadratic. For example, the two connected Wick contractions at the third leg pair its two fields with the first two legs, givingSumming the three placements provesThis is the leading local primordial bispectrum. The factor $6/5$ is the quadratic coefficient $3/5$ multiplied by the two cross-pairings. Terms of order $f_{\mathrm{NL}}^2$ have an odd Gaussian moment and vanish.

The literal quadratic local model also has a connected cubic-in-$f_{\mathrm{NL}}$ contribution, from one quadratic factor at every leg:This is the loop correction to the local primordial bispectrum; regulators may be needed for idealized spectra. Thus the PDF's displayed formula is the tree/leading-order result, not an exact identity for arbitrary $f_{\mathrm{NL}}$. The weak-non-Gaussian expansion assumes these loop terms are small. The local shape is enhanced in the squeezed bispectrum configuration when a long-wavelength perturbation modulates small-scale power.

Use the Fourier transform convention $\hat{w}(E)={\int }_{\mathbb{R}}w(t)e^{itE}dt$. Since $w\ge 0$, it is real, so $\hat{w}(-E)=\hat{w}(E{)}^{\ast }$. The positive-frequency cutoff therefore also eliminates frequencies at or below $-\Delta$. Normalization gives $\int w=\hat{w}(0)=1$.

For spectral filtering of Hamiltonian terms, chooseIn an energy eigenbasis, its matrix elements areAt $i=j=0$ the frequency is zero, so the normalization preserves the ground-state expectation:The spectral filter is a positive weighted average of unitary conjugations; in particular the integral is bounded in operator norm by $\Vert h_Z\Vert$. We can choose it even without an extra assumed tail bound: the evenization of a nonnegative bandlimited filter proved in part (e) produces another admissible spectral filter with the same type of positive-time decay. Make that choice consistently in all the filtered terms and shell definitions below.

Use the Fourier transform for the spatial Cauchy problem, or its periodic analogue, and regard the two starting levels as independently perturbed data. A Fourier mode with spatial factor $e^{im\theta }$ has amplification roots $G$ satisfyingPut $c=2\mu -1$ and $G=e^{i\theta /2}q$. ThenIf $|c|<1$, both roots have modulus one and their separation is bounded below uniformly in frequency:The uniform power bound from separated amplification roots now controls the two-level companion matrix for every time step. Its entries are uniformly bounded, and its eigenvector conditioning is bounded by the reciprocal root gap. The Parseval identity transfers this frequency-uniform bound to the spatial ${\ell }^2$ norm. This proves stability, rather than merely checking each root's modulus.

If $|c|>1$, the frequency $\theta =\pi$ has a root outside the unit disk, so there is exponential instability. If $|c|=1$, the two roots at $\theta =\pi$ coincide on the unit circle. The companion matrix is not a scalar matrix and has a nontrivial Jordan block; its powers grow linearly in the number of steps. Frequencies arbitrarily near that value produce the same lack of a uniform bound for localized Fourier packets, so this also invalidates Cauchy ${\ell }^2$ stability, not only periodic plane-wave stability. At $\mu =1$ the double amplification root is $-1$, and at $\mu =0$ it is $1$.

Therefore the full two-level stability range for a fixed positive Courant ratio isThe endpoint $\mu =0$ is moreover not a positive time step. Bounds deteriorate as $\mu$ approaches either endpoint; the displayed range is not a uniform claim over ratios arbitrarily close to one. A prescribed starter that removes one special parasitic component can change behavior for selected initial data, but does not establish the requested unrestricted two-level stability.

Use complex-linear distribution pairings, without conjugation. Write $\mathcal{D}=C_c^{\infty }({\mathbb{R}}^n)$ for the space of test functions. The smoothing convolution with a test function isOn a compact set of $x$ values, all the translated test functions have support in one compact set. Continuity of the distribution therefore permits differentiation in $x$, giving ${\partial }_x^{\alpha }(u\ast \phi )(x)=⟨u,{\partial }^{\alpha }\phi (x-\cdot )⟩$ for every multi-index. In particular this convolution is a smooth function, even if $u$ is not tempered.

For the first associativity identity, integration against $\psi$ and the distribution pairing can be interchanged: the integrand has a common compact support in $y$, depends smoothly on the integration variable, and satisfies the finite-order continuity estimate there. ConsequentlyThe common-support argument matters: a general distribution cannot be paired with arbitrary noncompact functions.

For convolution of distributions with a compactly supported factor, first take $v$ with compact support $K$ and defineThe inner pairing is interpreted using a cutoff function equal to one near $K$. It is smooth in $x$ and has support in $\mathrm{supp}\chi -K$. To see continuity, restrict $\chi$ to a fixed compact support $A$. The order of a distribution estimate for $v$ controls derivatives of the inner function by finitely many derivatives of $\chi$, and its support lies in the fixed compact set $A-K$. Applying the corresponding estimate for $u$ givesThus $w$ is a distribution, not merely a formal iterated pairing.

Choose an additional cutoff function in $x$ equal to one on a neighborhood of $A-K$. The resulting joint kernel is compactly supported in both variables, so the tensor product of distributions permits reversing the pairings. One justification is to approximate that smooth compact kernel, in all the required derivative seminorms, by finite sums of products of one-variable kernels; the two orders agree on such products and their continuity estimates pass to the limit. Reversing $x,y$ consequently gives $u\ast v=v\ast u$. If $u$ rather than $v$ has compact support, use the same construction with the roles reversed; pairing $u$ against a smooth function is then legitimate.

Evaluating the resulting smoothing convolution with a test function givesIf $v$ has compact support, $v\ast \phi$ is itself a test function; if $u$ has compact support, its action on the smooth inner convolution uses a cutoff. This explains the meaning of the formula in either case. It also proves uniqueness: $(w\ast \phi )(0)=⟨w,\phi (-\cdot )⟩$, and reflection runs through all test functions. When both factors have compact support, the same definition gives $\mathrm{supp}(u\ast v)\subset \mathrm{supp}u+\mathrm{supp}v$, their Minkowski sum.

The Schwartz space consists of smooth functions for which every seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }\varphi (x)|$ is finite. A tempered distribution is a continuous linear functional on this space. Fix the angular-frequency Fourier transform conventionThe Fourier transform isomorphism of the Schwartz space makes the dual definition continuous. For a compactly supported distribution, a fixed cutoff function $\chi =1$ near its support extends the action to smooth functions by $⟨u,g⟩=⟨u,\chi g⟩$. A finite-order estimate controls this by finitely many Schwartz space seminorms, so both compactly supported factors are tempered.

Their Fourier transform of a compactly supported distribution is the smooth function $\hat{u}(\lambda )=⟨u,e^{-i\lambda \cdot x}⟩$. Applying the compact-support convolution definition to the exponential givesThe convolution theorem has no extra factor with this normalization.

For the spherical surface measure convolution, put $k=|\lambda |$. Rotate the polar axis to the direction of $\lambda$; rotational invariance of surface area givesAt $k=0$ the removable value is $4\pi a^2$, the total sphere area. Hence $\widehat{u_a\ast u_b}(\lambda )=16{\pi }^{2}ab\sin (ak)\sin (bk)/k^2$.

For $r=|x|>0$, angular integration in Fourier inversion now givesThis conditional integral can be made rigorous by first inserting $e^{-\epsilon k}$ and then taking $\epsilon \downarrow 0$ in tempered distributions. The supplied sine identity gives an integral of $\pi /4$ when $|a-b|<r<a+b$, and zero off that interval. Thusas a regular distribution. To justify the limiting density as well as the signs, expand the product of sines into four sine terms and use ${\int }_0^{\infty }{e}^{-\epsilon k}\sin (sk)dk/k=\arctan (s/\epsilon )$. The four arctangents are uniformly bounded; the regularized inverse is bounded by a constant times $1/r$, which is a locally integrable function in three dimensions. Dominated convergence theorem therefore identifies the distributional limit with the displayed density.

Changing the two endpoint sphere values does not change the regular distribution; this includes the source's closed-interval representative. At a jump, symmetric Fourier inversion instead takes the half-value. If $a=b$, the $1/r$ singularity at the origin remains locally integrable and is not a point mass. As a normalization check,exactly the product of the original sphere areas.

Use $D_j=-i{\partial }_{x_j}$, so the Fourier transform of $D^{\alpha }u$ is ${\lambda }^{\alpha }\hat{u}$. Distinguish the full degree-$N$ polynomial $P(\lambda )$ from its homogeneous top-degree part $p_N(\lambda )$, the principal symbol. The elliptic differential operator condition isIt does not require the full polynomial to be nonzero at small frequencies. Compactness of the unit sphere gives $c_0={\min }_{|\theta |=1}|p_N(\theta )|>0$. Homogeneity and the lower-degree remainder implyFor sufficiently large $|\lambda |$ the second term is at most half the first, so the high-frequency lower bound for an elliptic polynomial isHere $⟨\lambda ⟩$ is the Japanese bracket. The order-zero case just means a nonzero constant and is immediate.

For real $s$, the Sobolev space definition with the current Fourier normalization isChanging the harmless constant in the norm gives the same space. A distribution on $X$ belongs to the Local Sobolev space $H_{\mathrm{loc}}^s(X)$ when $\chi u$, extended by zero outside $X$, belongs to $H^s({\mathbb{R}}^n)$ for every test function $\chi \in C_c^{\infty }(X)$.

A compactly supported distribution of finite order of a distribution $m$ has a smooth Fourier transform satisfying $|\hat{u}(\lambda )|\le C(1+|\lambda |{)}^m$, by applying the finite-order estimate to a fixed cutoff times the exponential. Thus its weighted squared transform is bounded by $C⟨\lambda {⟩}^{2(s+m)}$. This is integrable precisely in the sufficient range $2(s+m)<-n$, and provesThis is the negative Sobolev regularity of a compactly supported distribution; the strict inequality is important, and is not a claim that this sufficient threshold is optimal for each distribution.

To establish elliptic regularity without assuming the answer as an a priori smoothness hypothesis, first obtain a global constant-coefficient gain. For a compactly supported distribution $w$ with $P(D)w\in H^q$, the polynomial lower bound at high frequency givesOn $|\lambda |\le R$, the transform of $w$ is smooth and bounded. Therefore $P(D)w\in H^q$ implies $w\in H^{q+N}$. Possible low-frequency zeros of $P$ are harmless; we never divide by them.

Two elementary mapping facts supply the variable-coefficient argument. Distributional derivatives of order $j$ map $H^r$ into $H^{r-j}$. Sobolev multiplication by a smooth cutoff is bounded on $H^r$ for every real $r$, including negative ones. Indeed, for a compactly supported smooth $a$, the Peetre weight inequalityand $\widehat{aw}=(2\pi {)}^{-n}\hat{a}\ast \hat{w}$ reduce the bound to Young's convolution inequality with the integrable kernel $⟨\zeta {⟩}^{|r|}|\hat{a}(\zeta )|$. Smooth coefficients only need this property on compact subsets, where they can be multiplied by another cutoff.

Write the lower-order part as $B={\sum }_{|\alpha |<N}{f}_{\alpha }{D}^{\alpha }$, with $N\ge 1$, and suppose provisionally that $u\in H_{\mathrm{loc}}^r$ on a relatively compact neighborhood. For a cutoff function $\chi$ supported there,The bracket is the operator commutator. Its order is at most $N-1$: in the product rule, every surviving term has at least one derivative falling on $\chi$. Choose a second cutoff function equal to one near $\mathrm{supp}\chi$ when estimating the products. The derivative and smooth-multiplication bounds then giveThe global gain just proved applies to the compactly supported distribution $\chi u$, and yieldsThis is the cutoff bootstrap for local elliptic regularity; it works for real indices, not just nonnegative integers.

There is always a legitimate starting index. Around any fixed point choose ${\chi }_0=1$ on a smaller neighborhood. The compactly supported distribution ${\chi }_{0}u$ has the negative Sobolev regularity proved above, so $u\in H_{\mathrm{loc}}^{r_0}$ on that neighborhood for some finite $r_0$. The index need not be uniform over all of $X$. Repeating the one-step gain finitely many times reaches $s+N$, or the target is already reached if $r_0\ge s+N$. Since the point was arbitrary,For $N=0$ the result follows directly by dividing by the nonzero constant.

Finally, if $Lu=0$, its forcing belongs to every Local Sobolev space. The gain consequently gives every local Sobolev order for $u$. For each integer $j\ge 0$, choose an order larger than $j+n/2$ and apply the Sobolev embedding theorem to a localized solution. It has a $C^j$ representative; the representatives agree because they represent the same distribution. Thus every distributional solution of $Lu=0$ is smooth on $X$.

Work first with $q$ in the Schwartz space, so that all Fourier manipulations and spectral contour integrals are justified; weaker classes follow by the usual density or distribution arguments. DefineThe phase is purely imaginary. Since ${\partial }_{\bar{z}}E=kE$, setting $F=E\Phi$ reduces the spectral equation to ${\partial }_{\bar{z}}\Phi =E^{-1}q$. The whole-plane Cauchy-Pompeiu formula therefore constructs the solution decaying spatially at infinity:The freedom to add $E$ times an entire function is removed by this decay condition. For every fixed $k$, the integral is a spatial Cauchy-Green operator applied to a modulated source.

Now differentiate in the conjugate spectral parameter. The two exponential derivatives produce $\zeta -z$, canceling the Cauchy denominator, soThis is the spectral dbar equation: its right-hand side is the forward transform of $q$ multiplied by a known plane wave. Apply the whole-plane Cauchy-Pompeiu formula again, now in $k$:The spatial spectral equation also gives $F(z,k)=-q(z)/k+o(k^{-1})$ as $|k|\to \infty$. One way to justify this is to integrate by parts in the first Cauchy integral: $F=-q/k+T_k(q_{\bar{z}})/k$, where the modulated Cauchy integral $T_k(q_{\bar{z}})$ tends to zero by the Riemann-Lebesgue lemma. Comparing the $1/k$ coefficient of the spectral contour integral therefore givesThis derives the transform pair from two uses of the Cauchy-Pompeiu formula, not from an assumed inversion formula.

To identify the usual normalization, write $z=x+iy$ and $k=k_1+i{k}_2$. Then $\bar{k}z-k\bar{z}=2i(k_{1}y-k_{2}x)$. Set ${\xi }_1=2k_2$, ${\xi }_2=-2k_1$, so $d{\xi }_{1}d{\xi }_2=4dA(k)$. The result is exactly the two-dimensional Fourier transform pairThe factor four in the real-frequency change of variables is essential.

Fix the mixed Fourier transform conventionThe spatial derivative transforms to $2\pi ik\hat{f}$, and multiplication by $v$ transforms to $-(2\pi i{)}^{-1}{\partial }_{\xi }\hat{f}$. Hence the transformed free transport equation isIts characteristic equations for a transport equation give $\dot{\xi }=-k$, so the characteristic ending at $\xi$ at time $t$ began at $\xi +kt$. ConsequentlyThe same sign follows directly by substituting $x=y+tv$ in the Fourier transform of $f_0(x-tv,v)$. No first velocity moment is assumed, so the differential equation may be understood in the sense of tempered distributions; the explicit transform formula is valid pointwise because $f_0$ is integrable.

Retain $a=p_0^{1/\gamma }/{\rho }_0$ from part (b), and use time translation invariance to write $G(\mathbf{x},t;\mathbf{y},\tau )=g(\mathbf{x},\mathbf{y};t-\tau )$. Define the temporal Fourier transform byMultiplying the transformed Green function equation by $-a(\mathbf{x})$ givesThis is a symmetric divergence-form spatial operator, although the original unweighted operator need not be symmetric in the ordinary volume measure.

Let $G_j(\mathbf{x})=\tilde{G}(\mathbf{x},{\mathbf{y}}_j;\omega )$. The product rule gives the bilinear Green second identityThe frequency terms cancel. Integrating over the domain, the right side becomesThe boundary integral is zero for common homogeneous Dirichlet boundary condition, Neumann boundary condition or reciprocal Robin boundary condition conditions. In an unbounded domain, use the same outgoing limiting-absorption prescription for both Green functions; it gives the corresponding vanishing boundary pairing. This identity has no complex conjugation: it proves transpose wave reciprocity, not a Hermitian or time-reversal identity. We obtain the weighted acoustic Green-function reciprocityThe factor is frequency independent, so inverse Fourier transform gives the same relation at equal time lag. Both time arguments below have lag $t-{\tau }_2$, henceReciprocity exchanges source and receiver while preserving elapsed time; it does not turn a causal response into an advanced one.