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Set $\Delta =T/M$, $q_0=q$, and $q_M=q^{\prime }$. The Trotter product formula separates the kinetic and potential energy evolution to first order in $\Delta$. Inserting $M-1$ resolutions of the identity in the position operator basis and using the free short-time kernel givesThe exponent converges formally to the action along the path. The prefactors and intermediate integrations define the time-sliced configuration-space path integral measure, soA real-time path integral is an oscillatory limit with the Feynman i-epsilon prescription, rather than an ordinary probability integral. The operator derivation assumes the usual self-adjointness and product-formula hypotheses for the Hamiltonian.

For a quadratic potential energy write $q=q_c+\eta$, with $\eta (0)=\eta (T)=0$ and ${\ddot{q}}_c+V^{\prime }(q_c)=0$. The first variation vanishes by the Euler-Lagrange equation and the fixed endpoints. Because $V$ is quadratic, the remaining expansion is exact:Here integration by parts has no boundary contribution. Translation of the integration variables leaves the time-sliced measure unchanged. Thus the Gaussian path integral over $\eta$ is independent of the endpoint values, and the entire endpoint dependence lies in $e^{iS[q_c]}$.

Expanding $\eta$ in eigenfunctions satisfying Dirichlet boundary conditions, each real mode contributes an inverse square root of its quadratic eigenvalue, with the oscillatory phase fixed by continuation. Normalizing against $A_0={\partial }_t^2$ gives a precise determinant-ratio form:The determinant ratio is meaningful after a common regulator; the phase is part of the prescription. It is equivalent to writing $N=C(T)({\det }_{D}A{)}^{-1/2}$ with an endpoint-independent normalization.

For example, if $V^{\prime \prime }={\omega }^2$, the Dirichlet oscillator determinant ratio isA linear or constant term in $V$ changes the action but not this prefactor. The formula holds away from conjugate times, when the boundary problem and determinant are nonsingular. At $\sin \omega T=0$, it must be continued as a distribution with its Maslov index phase; the assertion of an ordinary finite prefactor cannot be used literally there. For the unshifted oscillator, at $T=r\pi /\omega$ the kernel is $e^{-ir\pi /2}\delta (q^{\prime }-(-1{)}^{r}q)$.

For the full-line source integral, impose vacuum boundary conditions by adiabatic damping. Take the source initially as a test function, so its pairings with the Green function are defined. After integration by parts, writeWith the Fourier transform convention $G(t)=\int d\omega e^{i\omega t}\tilde{G}(\omega )/(2\pi )$, the inverse isSet $q=\eta -GJ$. Completing the square givesand consequentlyThe Gaussian functional integral $Z[0]$ is source independent and contains the regulated determinant. This is most naturally read as a normalized vacuum generating functional. For $m>0$, closing the frequency contour gives $G(t)=-i{e}^{-im|t|}/(2m)$; the time-ordered oscillator two-point function is $iG(t)$. This checks the sign of the source exponent and identifies the required vacuum prescription.

Use $D_j=-i{\partial }_{x_j}$ and $⟨\xi ⟩=(1+|\xi {|}^2{)}^{1/2}$, so the Fourier transform of $P(D)u$ is $P(\xi )\hat{u}(\xi )$. Write $P_N$ for the homogeneous degree-$N$ part. An elliptic differential operator has $P_N(\xi )\ne 0$ for every real $\xi \ne 0$. By continuity on the unit sphere, $m={\min }_{|\omega |=1}|P_N(\omega )|>0$. The lower-degree terms are bounded by $C|\xi {|}^{N-1}$ for $|\xi |\ge 1$, whencefor sufficiently large $|\xi |$. This proves the high-frequency lower bound for an elliptic polynomial. Degree zero just means a nonzero constant and has no derivative gain.

With the Fourier convention in Question 3, the Sobolev space isThe condition includes that the weighted transform is represented by an $L^2$ function. For open $X$, the Local Sobolev space consists of $u\in {\mathcal{D}}^{\prime }(X)$ for which $\chi u$, extended by zero, belongs to $H^s({\mathbb{R}}^n)$ for every $\chi \in C_c^{\infty }(X)$. We use the elementary Sobolev multiplication by a smooth cutoff fact for every real $s$. It follows from Fourier convolution with the rapidly decreasing $\hat{\chi }$, the weighted inequality $⟨\xi {⟩}^s\lesssim ⟨\xi -\eta {⟩}^{|s|}⟨\eta {⟩}^s$, and the $L^1\ast L^2\to L^2$ convolution bound.

For a compactly supported distribution, continuity on test functions supported in a fixed compact neighborhood gives finite order: for some integer $M$,Insert a compact smooth cutoff function equal to one near the support, times $e^{-ix\cdot \xi }$. The resulting Fourier transform is smooth and satisfies $|\hat{u}(\xi )|\le C^{\prime }⟨\xi {⟩}^M$. The distribution is also tempered by this same finite-order bound. ThusIndeed the square of the weighted bound is integrable exactly when $2(t+M)<-n$. This is negative Sobolev regularity of a compactly supported distribution.

We next prove local regularity by a high-frequency reciprocal parametrix kernel, including its off-diagonal smoothing property. Choose $\psi \in C_c^{\infty }({\mathbb{R}}^n)$ equal to one on a ball containing all real zeros of $P$, and put $b(\xi )=(1-\psi (\xi ))/P(\xi )$, defining it smoothly as zero in the inner ball. Differentiation of the reciprocal and the elliptic lower bound giveThe Fourier multiplier $E=b(D)$ maps $H^s$ to $H^{s+N}$ by its zeroth-order bound, andThe kernel $K={\mathcal{F}}^{-1}b$ is smooth away from the origin: for $x\ne 0$, repeatedly integrate by parts using $e^{ix\cdot \xi }=(i|x{|}^2{)}^{-1}x\cdot {\partial }_{\xi }{e}^{ix\cdot \xi }$. After sufficiently many integrations, the differentiated symbol is integrable. For any desired $x$ derivative, repeat the argument with the additional polynomial ${\xi }^{\beta }$. A large-radius cutoff function justifies each step and its removal uniformly on compact sets away from zero. Thus convolution by $K$ carries a compactly supported distribution to a smooth function at points separated from its support. Also $R$ carries a compactly supported distribution to a Schwartz function, because $\psi \hat{u}$ is smooth with compact support.

For a target compact set in $X$, choose $\chi \in C_c^{\infty }(X)$ equal to one on a neighborhood of it. Then $\chi u$ and $[P(D),\chi ]u$ have compact support, and the commutator is supported where derivatives of $\chi$ occur, away from the target. The parametrix identity givesThe first term is in $H^{s+N}$ since $\chi P(D)u\in H^s$. The other two terms are smooth near the target by the proved kernel property. Since the target was arbitrary,This proof does not discard the cutoff function commutator; it places its support away from the set where regularity is sought.

For the final polynomial, select a multi-index ${\alpha }_0$ with $|{\alpha }_0|=N$ and ${\partial }^{{\alpha }_0}Q$ a nonzero constant. The derivative-ratio hypothesis immediately gives $|Q(\xi )|\ge c|\xi {|}^{\delta N}$ at large frequency. In particular $Q$ has no real zero there. For $N>0$, its degree bound also forces $\delta \le 1$. Differentiating $1/Q$ gives products of ratios ${\partial }^{\beta }Q/Q$, whose total derivative order is $|\alpha |$. Hence the corresponding high-frequency reciprocal satisfiesIts multiplier maps $H^s$ to $H^{s+\delta N}$. Its kernel is still smooth off zero: each frequency integration by parts now lowers the order by $\delta$, so more iterations may be needed, but $\delta >0$ supplies arbitrarily much decay. The same separated-support commutator identity applies without a change. The derivative-ratio Sobolev gain for a polynomial operator is thereforeSimply estimating the commutator by its differential order would lose this sharper gain; the off-diagonal kernel argument uses the full derivative-ratio hypothesis.

A Schwartz function is a smooth function for which every seminormis finite. These seminorms define the Fréchet topology of the Schwartz space $\mathcal{S}({\mathbb{R}}^n)$. The tempered distributions form its continuous linear dual ${\mathcal{S}}^{\prime }$. We use bilinear distribution pairing, with no complex conjugation on the test function.

Differentiation under the Fourier integral and integration by parts giveTherefore each Schwartz seminorm of $\hat{\phi }$ is bounded by a finite sum of $L^1$ norms of polynomially weighted derivatives of $\phi$. Insert the integrable weight $⟨x{⟩}^{-n-1}$ to bound those norms by finitely many Schwartz seminorms. This proves that $\mathcal{F}:\mathcal{S}\to \mathcal{S}$ is continuous.

To justify inversion without a merely formal exchange of integrals, insert $e^{-\epsilon |\xi {|}^2/2}$ in the inverse integral. The Gaussian Fourier integral givesThe normalized Gaussian is an approximate identity, so the right side tends to $\phi (x)$; the left side converges by dominated convergence theorem since $\hat{\phi }$ is integrable. It follows thatReflection is continuous in Schwartz seminorms, so ${\mathcal{F}}^{-1}=(2\pi {)}^{-n}\mathcal{R}\mathcal{F}$ is continuous as well. This proves the Fourier transform isomorphism of the Schwartz space.

Define the distributional Fourier transform byIt agrees with the ordinary integral transform whenever Fubini applies. Transposing the Schwartz inverse gives an inverse on ${\mathcal{S}}^{\prime }$, and the same squared-transform identity holds. For the usual strong dual topology, a seminorm is $p_B(u)={\sup }_{\phi \in B}|⟨u,\phi ⟩|$ for a bounded set $B\subset \mathcal{S}$. Since a continuous linear Schwartz map takes bounded sets to bounded sets, $p_B(\hat{u})=p_{\mathcal{F}B}(u)$ proves continuity, and similarly for the inverse. Continuity also holds in the weak dual topology. Thus the transform is a continuous isomorphism on tempered distributions, with the stated normalization.

For a real symmetric positive-definite matrix $G$, there is $c>0$ with $g(x)=x^{T}Gx\ge c|x{|}^2$. The reciprocal is locally integrable in dimension three: the radial factor near zero is $r^2{r}^{-2}dr$. At infinity, rapid decay of a Schwartz test function makes it integrable. More quantitatively,This single seminorm bound proves a tempered distribution. No principal-value extension is needed at the origin.

First calculate the isotropic transform. For $f_{\epsilon }(x)=e^{-\epsilon |x|}/|x{|}^2$ and $k=|\xi |>0$, spherical integration and the supplied sine-integral identity giveThe original functions converge in ${\mathcal{S}}^{\prime }$ to $|x{|}^{-2}$ by dominated convergence theorem against tests. The transforms are bounded by $2{\pi }^2/|\xi |$, which is locally integrable in three dimensions and integrable against Schwartz functions at infinity. HenceLet $y=G^{1/2}x$. The Jacobian is $(\det G{)}^{-1/2}$ and the dual vector is $G^{-1/2}\xi$. The Fourier transform of a reciprocal positive quadratic form is consequentlyThe frequency-origin value is understood distributionally; there is no additional delta term.

For the complex extension use a symmetric matrix $A=G+iB$ with real symmetric $B$ and positive real part $G$. Symmetry is natural for a quadratic form; a skew-symmetric part contributes nothing. The bound $|x^{T}Ax|\ge x^{T}Gx$ again gives a regular tempered reciprocal. Both sides of the prospective formula depend holomorphically on the matrix while its real part is positive: compact parameter sets give a common $|x{|}^{-2}$ bound for pairing and differentiated integrands. Continue from the real positive matrices along $A_z=G+izB$. In a complex neighborhood of $z=0$, imaginary $z$ gives real positive matrices, so the one-variable identity theorem supplies the equality; connected continuation along $0\le z\le 1$ reaches $A$.

The analytic determinant square root for accretive symmetric matrices must follow that continuation, rather than an arbitrary scalar principal root of $\det A$. Explicitly, put $C=G^{-1/2}B{G}^{-1/2}$ and let its real eigenvalues be $b_j$. Thenwhere each factor has positive real part. This is the determinant square root normalized positively on real positive matrices. AlsoThus the quadratic-form square root has the unambiguous branch with positive real part. The accretive complex quadratic reciprocal Fourier transform isUniform local integrability of the right side justifies its analytic continuation as a tempered distribution, not merely pointwise away from the origin.

For the particular form,Substitution gives the required normalized expressionFor every nonzero real $\xi$, the radicand has positive real part, so the specified square root exists uniquely. Both sides are regular tempered distributions, despite their locally integrable singularities at zero.

By the extreme value theorem, $p_0={\min }_{[-1,1]}p$ is strictly positive, and $p,q,r$ are bounded. The Cauchy-Schwarz inequality therefore givesFor completeness, the clamped interval derivative norm bound follows from $v^{\prime }(-1)=v(-1)=0$ and the one-dimensional Sobolev representative: integrate $v^{\prime \prime }$ to recover $v^{\prime }$, and then $v^{\prime }$ to recover $v$. Applying Cauchy-Schwarz and integrating over an interval of length two givesConsequently $\Vert v{\Vert }_{H^2}^2\le 21\Vert v^{\prime \prime }{\Vert }_2^2$. Nonnegativity of $q,r$ yields the precise coercive bilinear form boundThe Lax-Milgram theorem states that a bounded bilinear form on a real Hilbert space satisfying $a(v,v)\ge c\Vert v{\Vert }^2$ with $c>0$ represents every bounded linear functional uniquely: there is exactly one $u$ with $a(u,v)=\ell (v)$ for all $v$. All of its hypotheses have now been verified on $H$.

This weak solution is the unique global minimum, rather than merely a stationary point. The Ritz-Galerkin equivalence for a symmetric coercive form here is the exact identityIt is strictly positive for every nonzero $w\in H$. Conversely, a minimum has zero first variation in every direction, so it satisfies the same weak equation. Density of compactly supported test functions and boundedness of $a,\ell$ extend the distributional equation back to all $H$ test vectors. There is exactly one minimum, and it is exactly the clamped weak solution. No additional coefficient smoothness or classical fourth derivative is needed.

Let $I=(-1,1)$ and use the clamped second-order Sobolev spaceHere weak derivatives up to order two are square integrable, with Sobolev norm $\Vert v{\Vert }_{H^2}^2=\Vert v{\Vert }_2^2+\Vert v^{\prime }{\Vert }_2^2+\Vert v^{\prime \prime }{\Vert }_2^2$. The boundary values are the continuous traces given by the Sobolev trace theorem, not arbitrary pointwise representatives. This space is the $H^2$ closure of $C_c^{\infty }(I)$, so it is a closed Hilbert space. The derivative boundary conditions are essential, and $H_0^1(I)$ alone is insufficient.

Define the symmetric bilinear form and bounded linear functionalThen the variational problem isFor complex data replace $a$ by its sesquilinear version, take the real part of the forcing term and use absolute squares in $J$; the same argument applies.

Its first variation is $J^{\prime }(u)[v]=a(u,v)-\ell (v)$. Thus stationary points satisfy $a(u,v)=\ell (v)$ for every $v\in H$, the weak formulation of a clamped fourth-order equation. For smooth coefficients and functions, twice integrating the highest term and once the middle term gives the differential equation, with all boundary terms zero because $v=v^{\prime }=0$ at both ends. With only the stated continuous coefficients, this manipulation is instead understood distributionally: compactly supported test functions give $(p{u}^{\prime \prime }{)}^{\prime \prime }-(q{u}^{\prime }{)}^{\prime }+ru=f$ as a distribution. This avoids assuming nonexistent classical derivatives of $p$ or $q$.

Use the Fourier transform convention $\hat{F}(k)={\int }_{\mathbb{R}}F(x)e^{-ikx}dx$, with inverse factor $1/(2\pi )$. Splitting the inverse integral at zero givesThis is the Poisson kernel for the upper half-plane at height $a$.

Take the Fourier transform in $x$ of the Laplace equation. The transformed boundary-value problem isFor $k\ne 0$ the general solution is a linear combination of $e^{|k|y}$ and $e^{-|k|y}$; the two conditions set their coefficients to zero and one respectively. At $k=0$, the solution is the constant one. Inverting therefore givesThe upper boundary holds pointwise. At the lower boundary the Dirac delta function must be interpreted distributionally: the kernel has integral one and concentrates at zero, so for every continuous compactly supported test function $h$, $\int \psi (x,y)h(x)dx\to h(0)$. This also specifies the solution in the assumed class admitting the stated Fourier transforms.

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$: its elements are smooth functions with compact support. For a fixed compact set $K$, put ${\mathcal{D}}_K=\{\phi \in C^{\infty }(\mathbb{R}):\mathrm{supp}\phi \subset K\}$ and use the seminorms $p_m(\phi )={\max }_{0\le j\le m}\Vert {\phi }^{(j)}{\Vert }_{\infty }$. The space of test functions carries the usual test-function inductive limit topology of these spaces. In particular, a sequence converges precisely when its supports eventually lie in one compact set and every derivative converges uniformly. Thus ${\phi }_j\to \phi$ means that a common $K$ contains their supports and $p_m({\phi }_j-\phi )\to 0$ for every $m$.

A distribution is a continuous linear functional on this space of test functions, and ${\mathcal{D}}^{\prime }(\mathbb{R})$ denotes their space. We use complex-linear, bilinear pairings $⟨u,\phi ⟩$. Equivalently, for every compact set $K$ there are $C_K$ and a nonnegative integer $m_K$ such thatThe usual weak convergence of distributions is $u_j\to u$ if $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every test function. This specifies the convergence used below; it does not require convergence in any norm.

The distributional derivative is defined byThe derivative map sends ${\mathcal{D}}_K$ continuously to itself, with $p_m({\phi }^{\prime })\le p_{m+1}(\phi )$. The preceding continuity estimate therefore proves that $u^{\prime }$ is again a distribution. This definition extends the ordinary derivative of a smooth function, by integration by parts.

Choose the translation of a distribution convention ${\tau }_{h}f(x)=f(x-h)$. Its action on a test function isFor fixed $h$, the translated test functions have translated compact support and unchanged derivative sup norms, so this defines a distribution. For the differentiability of distribution translations, apply the Taylor theorem in integral form:All supports lie in one slightly enlarged compact set for $|h|\le 1$, and the same identity for every derivative proves uniform convergence. The continuity of $u$ now givesThe minus sign in the translation parameter is necessary for this convention.

For (a), choose a test function $\eta$ with $\int \eta =1$. If $\int \psi =0$, then $F(x)={\int }_{-\infty }^x\psi (s)ds$ is a test function, since the zero integral makes it vanish beyond both ends of the compact support. If $u_1^{\prime }=0$, then $⟨u_1,\psi ⟩=⟨u_1,F^{\prime }⟩=0$. Decompose $\phi =(\int \phi )\eta +\psi$ to obtainConversely, these constant regular distributions have zero distributional derivative. This also proves the general fact that a distribution with zero derivative is constant.

For (b), choose a cutoff function $\rho$ equal to one near zero. Every test function decomposes asThe quotient extends as a smooth function at zero, and it has compact support. If $x{u}_2=0$, the definition of multiplication of a distribution by a smooth function gives $⟨u_2,\phi ⟩=⟨u_2,\rho ⟩\phi (0)$. Conversely, $x{\delta }_0=0$. Thus the kernel of multiplication by a coordinate is exactlyIn particular, derivatives of the Dirac delta distribution are not additional solutions: $x{\delta }_0^{\prime }=-{\delta }_0$.

Now write $D=d/dx$ and $P(z)=-z^n+{\sum }_{j=0}^{n-1}{a}_j{z}^j$, where $n\ge 1$. The characteristic roots of a constant-coefficient differential equation are the distinct roots of a polynomial ${\lambda }_1,\dots ,{\lambda }_r$ of $P$, with multiplicities $m_1,\dots ,m_r$. The distributional regularity of a constant-coefficient ordinary differential equation can be proved without assuming regularity in advance. Put $q_j(z)=(z-{\lambda }_j{)}^{m_j}$ and $Q={\prod }_j{q}_j=-P$. Since the $q_j$ are coprime polynomials, polynomial division and Bezout identity give polynomials $b_j$ such thatFor example, invert $Q/q_j$ modulo $q_j$, sum the resulting expressions, and absorb a remaining multiple of $Q$ into one coefficient. For $P(D)v=0$, the kernel decomposition for coprime polynomials therefore givesThe Leibniz rule for multiplication of a distribution by a smooth function implies $D^{m_j}(e^{-{\lambda }_{j}x}{v}_j)=0$. Repeatedly using a distribution with zero derivative is constant shows that a distribution with $m$th derivative zero is a polynomial of degree at most $m-1$: subtract the polynomial primitive of its constant $(m-1)$st derivative, and induct on $m$. Consequently the most general exponential polynomial solution of a constant-coefficient differential equation isEvery displayed term is annihilated by $P(D)$, so all coefficients are allowed. For the linear independence of these $n$ functions, on a relation, apply ${\prod }_{i\ne j}(D-{\lambda }_i{)}^{m_i}$; on $e^{{\lambda }_{j}x}$ times a polynomial of degree less than $m_j$, each remaining factor acts invertibly on that polynomial space. Hence the $j$th polynomial must vanish. Every distributional solution is an analytic function, and in particular a classical solution. For real coefficients and real-valued distributions, take complex conjugate coefficients at conjugate characteristic roots of a constant-coefficient differential equation, or equivalently use the corresponding real sine and cosine forms.

For the last equation, the operator is $xP(D)$: the coordinate multiplies the result of differentiation. The kernel of multiplication by a coordinate says precisely thatChoose the retarded fundamental solution of a constant-coefficient ordinary differential operator $E=Hw$, where $H$ is the Heaviside function and $w$ is the analytic solution of $P(D)w=0$ withSuch $w$ exists uniquely by the elementary initial value problem for a constant-coefficient ordinary differential equation; alternatively the residue construction in the next solution gives it explicitly. The distributional jump formula for a Heaviside product isIt follows by induction from $D(Hw)=H{w}^{\prime }+w(0){\delta }_0$, using integration by parts. With the chosen initial derivatives, every lower-order jump term vanishes and the leading coefficient $-1$ gives $P(D)E={\delta }_0$. Subtracting $CE$ reduces the last equation to the homogeneous one. Thus the coordinate-degenerate constant-coefficient differential equation has exactly the solutionsThere are $n+1$ independent constants. For $n\ge 2$, derivatives through order $n-2$ match at zero, while the derivative of order $n-1$ may jump; for $n=1$, the function itself may jump. An arbitrary distribution concentrated at zero cannot be added, because its image under the nonzero leading derivative would contain a nonvanishing highest derivative of the Dirac delta distribution.

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 $-1$.

An oscillatory integral defines a distribution by cancellation, even when its oscillatory integral amplitude is not integrable in frequency. We use symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)=S_{1,0}^N(X\times {\mathbb{R}}^k)$, where $X\subset {\mathbb{R}}^d$ is open and $N\in \mathbb{R}$. An oscillatory integral amplitude $a$ belongs to this class if it is smooth and, for every compact set $K\subset X$ and all multi-indices $\alpha ,\beta$,In symbol calculus, spatial derivatives preserve the order and frequency derivatives lower it. The fixed number $N$ is independent of $K$; this will produce a finite global order of a distribution, although continuity constants may depend on $K$.