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The Malgrange–Ehrenpreis theorem states that every nonzero constant-coefficient linear differential operator on Euclidean space has a distributional fundamental solution. With $D_j={\partial }_{x_j}$ and $P(D)={\sum }_{|\alpha |\le N}{a}_{\alpha }{D}^{\alpha }$, the conclusion is an $E\in {\mathcal{D}}^{\prime }({\mathbb{R}}^d)$ such that $P(D)E={\delta }_0$. If $N=0$, simply take $E=a_0^{-1}{\delta }_0$; hence assume $N\ge 1$.

We construct a Hörmander staircase in frequency space. The highest-degree homogeneous part $P_N$ is not identically zero on real vectors, since a polynomial vanishing on all real vectors has all coefficients zero. After an orthogonal transformation of coordinates, we can arrange that $P_N(e_1)\ne 0$. Thus, for each real ${\xi }^{\prime }\in {\mathbb{R}}^{d-1}$, the polynomialhas degree $N$ in $z$ and the same nonzero leading coefficient $b=i^N{P}_N(e_1)$, independent of ${\xi }^{\prime }$. This constant leading coefficient of a polynomial is what makes a uniform staircase possible.

For a fixed ${\xi }^{\prime }$, factor $p(z,{\xi }^{\prime })=b{\prod }_{\nu =1}^N(z-z_{\nu })$, counting repeated roots of a polynomial. Consider the $N+1$ heights $h_j=3j$, $j=0,\dots ,N$. A given root has imaginary part within distance less than one of at most one of these heights. By the pigeonhole principle, some height avoids every root, and thenThis is finite-height polynomial root avoidance. Root labels need not be chosen continuously or even measurably. Instead define the closed setsContinuity in $s$ extends the inequality from rational $q$ to every real $s$. These Borel sets partition ${\mathbb{R}}^{d-1}$. The Hörmander staircase assigns the horizontal contour $s+i{h}_j$ over $A_j$; its height is bounded by $3N$, and the denominator has the uniform lower bound $|b|$. For $d=1$, the transverse space is a single point and only one horizontal contour is needed.

Use the Fourier transform convention $\hat{\phi }(\zeta )=\int e^{-ix\cdot \zeta }\phi (x)dx$, with inverse factor $(2\pi {)}^{-d}$. Define the candidate fundamental solution of a linear differential operator byFor a test function supported in a fixed compact set $K$, Fourier decay in a bounded complex strip gives, for any integer $M$,Indeed, this is the real Fourier transform of $e^{-h_j{x}_1}\phi (x)$ with reversed frequency, and repeated integration by parts with $(1-\Delta {)}^M$ proves the estimate. Taking $2M>d$ and using the denominator bound proves absolute convergence and a continuity estimate on ${\mathcal{D}}_K$. Thus $E$ is a distribution; no unsupported interpretation of a divergent inverse Fourier transform is being used.

For its distributional derivatives, the formal transpose of a differential operator is $P(-D)$. Sinceapplying $P(D)$ cancels the denominator. For each fixed ${\xi }^{\prime }$, the numerator is an entire function of $z$. The Cauchy integral theorem, applied to a rectangle between the lines $\mathrm{Im}z=0$ and $\mathrm{Im}z=h_j$, givesThe two vertical edges tend to zero by the same bounded-strip decay. The resulting contour deformation is performed separately for each transverse frequency, so discontinuities of the staircase height introduce no additional boundary terms. Absolute convergence permits integration over the partition $A_j$. The Fourier inversion theorem then yieldsThis proves the Malgrange–Ehrenpreis theorem. Undoing the orthogonal change of coordinates gives the fundamental solution of a linear differential operator for the original operator; the Dirac delta distribution is unchanged by that change of coordinates.

For a one-dimensional operator $L=P(D)$ of degree $N\ge 1$, with leading coefficient $a_N$, we may choose the single staircase contour below every pole. Take $\gamma >\max \{\mathrm{Re}\lambda :P(\lambda )=0\}$ and integrate on $\mathrm{Im}\zeta =-\gamma$. The corresponding formula is the Bromwich contour version of the construction above, after putting $\lambda =i\zeta$. Its poles all lie above the frequency contour. For $x<0$, close that contour downwards; the exponential function $e^{ix\zeta }$ decays and there are no enclosed poles. For $x>0$, close upwards, where the exponential function again decays, and apply the residue theorem. The bound $1/P(i\zeta )=O(|\zeta {|}^{-N})$ justifies the large arcs, including $N=1$ by the Jordan lemma away from their endpoints. Thus away from $x=0$ the retarded fundamental solution is $H(x)u(x)$, whereThe sum is over distinct roots of a polynomial, with the residue including the whole multiplicity. For simple characteristic roots of a constant-coefficient differential equation, it reduces to $u(x)={\sum }_{\lambda }{e}^{\lambda x}/P^{\prime }(\lambda )$. Repeated roots give exponential polynomial solutions of a constant-coefficient differential equation through differentiation of $e^{zx}$.

To establish the equality also at the origin, rather than leave a possible point-supported term undecided, verify the distributional jump formula for a Heaviside product. The residue expression is an entire function of $x$, andIts initial derivatives areThese identities follow by integrating over a large circle: for $j\le N-2$ the integrand is $O(|z{|}^{-2})$, whereas for $j=N-1$ its coefficient of $z^{-1}$ is $a_N^{-1}$. The distributional jump formula for a Heaviside product therefore gives $L(Hu)={\delta }_0$. Moreover, since $e^{-\gamma x}Hu$ is a tempered distribution, its Fourier transform satisfiesThere are no real zeros of this polynomial, so its reciprocal is the transform. This is exactly the shifted-contour construction, proving equality there as a distribution as well. HenceThe sign of $a_N$ matters: for the operator in the preceding solution, $a_N=-1$ and the last initial derivative is $-1$.

The bounded classical solution is the heat-kernel convolutionThe two-dimensional heat kernel has integral one. Boundedness of $g$ permits differentiation under the integral for $t>0$, proving $u_t=\Delta u$ and $|u|\le \Vert g{\Vert }_{\infty }$. Its approximate identity property gives $u(x,t)\to g(x)$, locally uniformly since $g$ is continuous. This bounded solution satisfies the required Gaussian-growth bound.

For heat-equation uniqueness under Gaussian growth, let $w$ be the difference of two solutions, so $w(x,0)=0$ and $|w|\le M_0{e}^{a|x{|}^2}$ on a finite time interval. Choose $b>a$ and a slab $0\le t\le \tau <1/(4b)$. The positive comparison solutionsatisfies $({\Phi }_b{)}_t=\Delta {\Phi }_b$. For any $\epsilon >0$, its faster spatial growth makes $|w|\le \epsilon {\Phi }_b$ on a sufficiently large lateral cylinder boundary. At the initial boundary, $w=0$. The heat equation maximum principle applied to $\pm w-\epsilon {\Phi }_b$ proves the same inequality inside. First allow the cylinder radius to increase, then let $\epsilon \downarrow 0$. Repeating these slabs proves uniqueness on every finite interval with the stated uniform growth bound. No spatial integrability of $g$ is required.

At time $T$, this is Gaussian filtering with covariance $2TI$, hence $\sigma =\sqrt{2T}$ per coordinate. With the angular-frequency Fourier transform convention $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$, the Gaussian filtering Fourier multiplier isFor merely bounded $g$, this identity is understood through tempered distributions. The multiplier leaves the zero frequency unchanged and attenuates large frequencies exponentially: it suppresses rapid noise fluctuations but blurs image edges. In cycles-per-length frequency $k$, the multiplier is $e^{-4{\pi }^{2}T|k{|}^2}$.

For the printed Perona-Malik equation, retain the supplied conductance $c(s)=s{e}^{-s^2/(2{\lambda }^2)}$, which includes an extra factor $s$. In one dimension write $p=u_x$ and $F(p)=c(|p|)p=p|p|e^{-p^2/(2{\lambda }^2)}$. Then $u_t=F^{\prime }(u_x)u_{xx}$, withThe forward-backward threshold for gradient-weighted exponential diffusion is thereforeAt $|u_x|=0$ and $\sqrt{2}\lambda$ the coefficient vanishes, giving degenerate diffusion. Increasing $\lambda$ increases the forward-diffusion range. Small nonzero slopes smooth, while large slopes formally sharpen; a negative coefficient produces short-wave growth and ill-posedness, so this is a dynamics explanation, not a general existence theorem for arbitrary data. Replacing the printed conductance by $e^{-s^2/(2{\lambda }^2)}$ would give threshold $\lambda$, but that is a different equation.

The four-neighbour mean expansion follows by Taylor expansion: opposing first and third derivatives cancel, soIf $u(x_0)={\mathrm{mean}}_h(u)(x_0)$ for every sufficiently small $h$, division by $h^2$ and passage to the limit give $\Delta u(x_0)=0$. This is a pointwise conclusion; it does not assert harmonicity in a whole neighbourhood.

For the area median over the disk, the disk-median curvature expansion, at a point where $|\nabla u|\ne 0$, isConsequently the analogous median fixed-point condition yieldsThis is the vanishing of the level-line curvature at $x_0$: to second order the level line is straight there. It need not be a straight segment. For example $u(x,y)=y+x^3$ has zero disk median at the origin for every radius, by odd symmetry, but its zero level line is the cubic $y=-x^3$. Vanishing curvature on an entire connected regular level arc would force that arc to be straight. The factor $1/6$ is for the disk's uniform area measure; a circle-boundary median has a different scale factor, while producing the same zero-curvature equation.

Nonnegativity and unit mass give, by the triangle inequality for the Fourier transform, $|\hat{f}(\xi )|\le \int f=1=\hat{f}(0)$. The finite second moment implies a finite first absolute moment, so differentiating under the Fourier integral is justified andThe exponent convention has no $2\pi$, so no such factor occurs in this derivative.

Apply the spherical identity of part (b) with $x=v-v_{\ast }$, $y=\xi$ and $\varphi (s)=e^{-is/2}$. The angular exponential in part (c) can be replaced, after angular integration, by $e^{-i|\xi |(v-v_{\ast })\cdot \sigma /2}$. The full phase is thenThe Fubini theorem factors the two velocity integrals into Fourier transforms. With ${\xi }^{\pm }=(\xi \pm |\xi |\sigma )/2$, the gain integral is exactly ${\int }_{{\mathbb{S}}^2}\hat{f}({\xi }^{+})\hat{f}({\xi }^{-})d\sigma$.

The loss Fourier transform is $\hat{f}(\xi )\hat{f}(0)$. Dividing the gain by the sphere area gives the Bobylev identity for the Maxwell molecule collision operator:The initial Fourier datum is ${\hat{f}}_0$. The surface measure on a sphere here is two-dimensional surface area, not the restriction of ambient three-dimensional Lebesgue measure, which would give the sphere measure zero.

Expansion of the two squares givesLet $D=d(f,g)$ be the Fourier distance of order two, with the supremum taken over $\xi \ne 0$. Both Fourier transforms have modulus at most one. Add and subtract $\hat{g}({\xi }^{+})\hat{f}({\xi }^{-})$, then use the triangle inequality:Dividing by $|\xi {|}^2$ proves the bound by $D$, and splitting the last expression into its two weighted terms gives the requested intermediate inequality. If one of ${\xi }^{\pm }$ vanishes, its unweighted difference is zero by equal mass; its weighted term is interpreted as zero, avoiding a $0/0$ quotient.

Equal mass and first moment also explain finiteness of this distance: subtract the constant and linear Taylor terms in the Fourier integral and use $|e^{-ia}-1+ia|\le a^2/2$. This bounds the difference by $|\xi {|}^2\int |v{|}^2(f+g)/2$. No direction-independent extension of the quotient at zero is required.

The ordinary Fourier transform does not exist. There is a real simple pole at $x=-1$, withThe two separate improper integrals diverge logarithmically. Thus no ordinary large-$k$ expansion is defined under the stated integral convention. A Cauchy principal value or a contour prescription would be additional data, and must be stated rather than silently introduced.

For completeness, the symmetric Cauchy principal value has a precise answer. The real pole contributes the principal-value Fourier transform of a real poleThe other poles are $z_{\pm }=(1\pm i\sqrt{3})/2$ with residues $r_{\pm }=1/(3z_{\pm }^2)=(-1\mp i\sqrt{3})/6$. For $k>0$, close the contour integration in the lower half-plane, with clockwise orientation. The nonreal pole contributes $-2\pi i{r}_{-}{e}^{-ik{z}_{-}}$, while the real principal-value pole supplies the half-residue above. For $k<0$, the upper-half-plane contour gives the conjugate result. With $s=\mathrm{sgn}(k)$, the exact principal-value formula for $k\ne 0$ isThese are its two nonzero asymptotic contributions: a nondecaying oscillatory real-pole term and an exponentially small complex-pole term. There is no second nonzero inverse-power term. This formula is qualified by the principal-value choice; it is not the ordinary transform requested in the statement.

Other prescriptions change the leading term. For example, replacing the real pole by its upper or lower boundary value changes the distribution by $\mp i\pi \delta (x+1)/3$, and hence changes the transform by $\mp i\pi e^{ik}/3$. This demonstrates why a pole prescription is essential. The smooth Taylor series at the origin alone would miss the decisive real-pole contribution.

Use the standard Fourier transform on the whole real axis. The printed lower limit is missing a minus sign. For this integrable even function, its only obstruction to smoothness is the origin, so its high-frequency algebraic terms come from its local cusp:Localize near zero with a smooth cutoff. The smooth polynomial terms, including $|x{|}^6=x^6$, contribute no algebraic cusp term; their localized transforms decay faster than any prescribed inverse power. The supplied half-line formula, interpreted away from $k=0$ in an Abel-regularized or tempered distribution sense, gives the Fourier transform of an algebraic cuspFor $p=3$ the cosine is $1$, and for $p=9$ it is $-1$. Both cusp coefficients in the local expansion are $-1$. HenceThe second coefficient is $29!=725760$. The absence of a $|k{|}^{-7}$ term follows from the smooth even power $x^6$, not from neglecting an available correction.

One can verify the signs by integration by parts on $2\mathrm{Re}{\int }_0^{\infty }{e}^{-ikx}(1+x^3{)}^{-1}dx$. Its endpoint derivatives first have nonzero relevant values $f^{(3)}(0)=-3!$ and $f^{(9)}(0)=-9!$, which contribute twice those values divided by $(ik{)}^4$ and $(ik{)}^{10}$. Higher derivatives are integrable on the half-line, justifying the displayed remainder after further integrations. Equivalently these coefficients are the Fourier decay from a derivative jump at the origin.

Use the Fourier transform pair $\hat{q}(k)=\int q(x)e^{ikx}dx$, $q(x)=(2\pi {)}^{-1}\int \hat{q}(k)e^{-ikx}dk$. The point force transforms to $F$. With the dispersion relation $D$ defined above, the sheet equation becomesThe point-force radiation from a fluid-loaded sheet is therefore represented exactly byThe causal contour and outgoing acoustic square-root branch are fixed first with $\mathrm{Im}\omega <0$, then continued to the desired real frequency. This prescription fixes how poles and the branch points are passed.

For the acoustic far field $x=r\cos \theta$, $y=r\sin \theta$, take $0<\theta <\pi$ bounded away from grazing and $k_{0}r\gg 1$. The method of steepest descent saddle point is $k_s=k_0\cos \theta$, with ${\gamma }_s=i{k}_0\sin \theta$. The supplied saddle point rule, including its $\sin \theta$ factor, gives, provided the contour deformation crosses no poles,A convenient simplification, free of division by $T$, isEquivalently, when $T\ne 0$,The expression printed in the PDF is missing sound-speed factors for general dimensional $c_0$. It agrees with this result if $c_0=1$ in fully normalized units; when $c_0$ is retained as an arbitrary sound speed, the numerator needs $c_0^{-3/2}$ and the structural term needs $T/c_0^2$ in the last form. These factors arise respectively from cylindrical spreading, the pressure-density relation, and $k_s=\omega \cos \theta /c_0$.

A direct countercheck is the transparent-sheet limit $m=T=0$. The sheet jump condition then gives ${\hat{p}}_{+}(k,0)=F/2$, so the saddle point rule requiresThe printed expression, interpreted continuously after multiplying out its structural factor, instead gives $F(2c_0{)}^{-1}\sqrt{\omega /(2\pi r)}\sin \theta$ times the same phase. It differs by a factor $c_0^{3/2}$; for example it is eight times too large when $c_0=4$. This limit also verifies the normalization of the corrected density field independently of the sheet's elastic-sheet tension.

To decide about poles, track the roots of $\Delta (k,\omega )$ on the chosen square-root sheet and deform the original causal contour to the steepest descent contour. A root contributes a residue exactly when it lies in the region swept out by that deformation; its sign is fixed by the contour orientation. Branch cuts must be retained throughout this comparison. Which roots are crossed can depend on observation angle, producing a change of the modal contribution when a pole meets the deformation boundary. A saddle point approaching a pole or a grazing endpoint requires an approximation uniform in that limit, rather than the isolated saddle point formula above.

The crossed poles are the free fluid-sheet modes of the preceding solution. Real subsonic roots represent evanescent acoustic surface waves carrying energy along the sheet, with normal decay; complex continuations represent leaky or radiating modes. Their residues must be added to the saddle point sound when the causal contour selects them. The specification “no poles contribute” is therefore a substantive condition on the contour, not permission to ignore zeros of the dispersion relation.

Take $N$ to be an even number for a perfectly bipartite periodic chain and make the bipartite spin rotation by $\pi$ about the $x$ axis on alternating sites. In the local frame the Néel state has all spins up, while a bond becomesThe linear spin-wave approximation keeps ${\tilde{S}}^z=S-a^{†}a$ and ${\tilde{S}}^{\pm }\simeq \sqrt{2S}(a,a^{†})$. ThereforeEvery site has two neighbours. Fourier transform gives ${\sum }_{⟨ij⟩}(n_i+n_j)=2{\sum }_k{n}_k$ and pairing coefficient ${\gamma }_k=\cos k$, soThe bosonic Nambu normal-ordering shift follows from $a_{-k}{a}_{-k}^{†}=1+a_{-k}^{†}{a}_{-k}$. It adds $NJS$ inside the matrix expression, which must be subtracted in its constant. HenceA periodic chain with an odd number of sites is frustrated at the boundary and lacks this exact two-sublattice reference; the bulk thermodynamic calculation uses the even-chain sequence.

Use the Fourier transform convention $\hat{\nu }(\xi )=\int e^{-ix\cdot \xi }d\nu (x)$ throughout this question. The decay assumption implies square integrability, since polar coordinates giveNear zero the integrand is bounded by $r$, and at infinity it is bounded by $r^{-1-2ϵ}$. The positive $ϵ$ is what makes the latter integrable.

By the Plancherel theorem, there is a function $g\in L^2({\mathbb{R}}^2)$ whose Fourier transform is $\hat{\mu }$. It is the density of $\mu$ with respect to Lebesgue measure. To justify this step rather than assume a density, for every Schwartz function $\phi$, Fourier inversion givesThe finite measure and the locally integrable function thus define the same tempered distribution, so they agree as measures: $d\mu =gdx$. In particular $g\ge 0$ almost everywhere and $\int g=\mu ({\mathbb{R}}^2)<\infty$. This is the L2 density from a square-integrable Fourier transform principle.

For $f\in L^{\infty }(\mu )$, the density of $fd\mu$ is $fg$. The inequality $|f|\le \Vert f{\Vert }_{L^{\infty }(\mu )}$ holds wherever $g>0$, apart from a Lebesgue measure zero set, soA second application of the Plancherel theorem yields the explicit boundThe implicit constant in the requested estimate may depend on the measure and its Fourier-decay bound, but is independent of $f$.

Parametrize the unit circle by $\omega (\theta )=(\cos \theta ,\sin \theta )$, with $-\pi \le \theta <\pi$ and $d\sigma =d\theta /(2\pi )$. Extract the constant phase at $\omega (0)$:On the support, $|\theta |\le 2\delta /C$. The frequency rectangle and the elementary bounds $|1-\cos \theta |\le {\theta }^2/2$, $|\sin \theta |\le |\theta |$ implyChoose the fixed constant $C$ large enough that this is less than $\pi /3$. Every phase then has real part at least $1/2$. Nonnegativity of $\psi$ prevents cancellation after this phase rotation, while its central plateau gives $\int \psi d\sigma \ge \delta /(\pi C)$. Hence the Fourier transform satisfiesThis is the circle cap Fourier lower bound. No upper bound on the values of $\psi$ is needed here; the support, plateau and nonnegativity suffice. The long radial scale ${\delta }^{-2}$ comes from the quadratic term $\cos \theta -1$, whereas the transverse scale ${\delta }^{-1}$ comes from the linear term $\sin \theta$.

Take the Fourier transform in $x$. The transform of the source is $\delta (t-t^{\prime })e^{-ik{x}^{\prime }}$, so the Green function equation becomesChoose the causal Green function, zero for $t<t^{\prime }$. Away from $t^{\prime }$ the solution is a decaying exponential, and integrating across $t^{\prime }$ gives the jump $\tilde{G}(t^{\prime }+)-\tilde{G}(t^{\prime }-)=e^{-ik{x}^{\prime }}$. ThereforeThe phase shifts the kernel's centre to $x^{\prime }$. Applying the inverse Fourier transform and the Fourier transform of a Gaussian with $a=D(t-t^{\prime })$ givesFor positive elapsed time this heat kernel solves the homogeneous diffusion equation and has unit integral; as elapsed time tends to zero it approaches the spatial delta distribution, supplying the required impulse.

Choose the Fourier transform conventionThen ${\partial }_a$ becomes $i{q}_a$. Applying this to the Lorenz constraint in Proca theory yieldsThus every physical Fourier transform mode is orthogonal to its four-momentum in the Minkowski metric.

There are two closely related objects to distinguish. Inverting the quadratic Proca action gives the usual covariant Proca propagator. After integration by parts, its kernel isThe matrix inverse with the Feynman i-epsilon prescription givesMultiplication by $K^{ab}$ gives $i{\delta }_b^a$ in the distributional limit. The numerator agrees with the polarization sum for a massive vector boson at the poles, but is not a transverse linear projection at arbitrary four-momentum.

For the literal canonical time-ordered product of $A_a$ and $A_b$, the nondynamical component $A_0$ produces the Proca time-ordering contact term. In the chosen time coordinate the full answer isTo see the local term directly, the three physical polarization vectors give ${\sum }_{\lambda }{ϵ}_{\lambda 0}^2=|\mathbf{q}{|}^2/m^2$. The canonical $00$ two-point correlation function therefore has Fourier transformBy contrast, ${\eta }_{00}+q_0^2/m^2=|\mathbf{q}{|}^2/m^2-(q^2+m^2)/m^2$, so the covariant expression contains an additional $i/m^2$. The mixed and spatial components have no additional contact term. ThusIf $T$ is used to mean covariant time ordering, commonly denoted $T^{\ast }$, the conventional answer is instead just ${\Delta }^{\mathrm{cov}}$. Both conventions have the same propagating poles and agree away from coincidence; explicitly separating them respects the printed definition as an ordinary time-ordered product.

Keep the mostly-plus Minkowski metric and the Fourier transform $e^{ip\cdot x}$. Write $/p={\gamma }^a{p}_a$, with $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}$. To fix the otherwise unspecified phase of ${\gamma }^5$ and the Dirac adjoint, takewhere ${\Gamma }^a$ are ordinary mostly-minus gamma matrices. Then $({\gamma }^5{)}^2=-1$, $\{{\gamma }^5,{\gamma }^a\}=0$, and the fermionic time-derivative term is $i{\psi }^{†}{\partial }_t\psi$. In these conventions the interaction with real $g$ is Hermitian: in mostly-minus notation it is $ig{\bar{\psi }}_{\mathrm{std}}{\Gamma }^5\psi \varphi$. If one instead calls $i{\gamma }^5$ the square-one chirality matrix, its coefficient must be $-ig$ to represent the same interaction. These phase choices leave physical relativistic scattering cross-sections unchanged.

Expanding $e^{i{S}_{\mathrm{int}}}$ yields the following Feynman rules with relativistically normalized external states:

Strip the overall four-momentum conservation delta function when defining the scattering amplitude. There are no further bare interaction vertices, no gauge fixing and no Faddeev-Popov ghost fields in this theory. Renormalized higher-order calculations add the required counterterms; these are additional to the rules of the displayed classical Lagrangian density.