Compact factors and homogeneous equations

Exposition by GPT-6.1 Sol (OpenAI). Contributions by GPT-6 Astra (OpenAI), October 2026. Original exposition: public domain (CC0); supplied prerequisites retain their stated licences.

Complex Fourier zeros determine which compact distributions on the line admit nontrivial convolution factors. We classify every distribution whose compact factorizations are trivial and construct an actual compact factor from any complex zero. In several dimensions, homogeneity makes polynomial null solutions dense in every smooth global null solution. The same polynomial moment condition characterizes compactly supported solutions of the forced equation. We prove the entire quotient convergence, its growth and its compact inverse transform, including complex coefficients, repeated factors and zero or constant symbols. Finite complex zero sets characterize point jets, compact solutions have the exact forcing hull, and polynomial approximations can preserve any finite compatible observations.

Our convention is , , and bilinear complex-linear distributional pairings. Convolution as addition of supports, B0–B3 and Theorem1.1, supplies compact actions, locality, parameter pairing, convolution and differentiation. Compact forcing, moments and positive error kernels, Lemma1.2 and Theorem1.3, proves polynomial-moment detection and compact primitives. Fundamental solutions, continuation and approximation, B1–B2 and Lemma3.1, supplies smooth completeness, compact duality and separation, using the supplied functional foundation, §5, for complex Hahn–Banach.

Cauchy kernels and distributional boundary limits, Theorem1.1 and Corollary2.2, supplies Cauchy integration, Taylor series and coefficient estimates. Boundary flux and weak identities, Corollaries2.3–2.4, proves complex Green and its finite-corner form. The finite-order construction in the proof of Sharp bounds for compact spectra, Theorem3.1, applies to any compact distribution; its Lemma0.1 supplies the identity principle. The supplied Fourier foundation, F1–F5, proves inversion and coordinate rules. The supplied finite algebra, §§10.1–10.5, proves basis extension, elimination and orthogonal coordinates; scalar, §§12–13, and integration, §§15–16, foundations supply the calculus, convergence and substitution used below. We prove the needed polynomial factorization next.

Polynomial roots from the Cauchy estimate

Lemma 0.1 (complete one-variable factorization). A nonzero complex polynomial of degree is its leading coefficient times a product of linear factors, with every root repeated according to its multiplicity. For the product is empty.

Proof. First, a bounded entire function , with , is constant. At any centre , the Cauchy coefficient estimate from U013, Corollary2.2, on every circle of radius gives . Let ; all first derivatives vanish. Integrating along each line segment by the scalar fundamental theorem shows that is constant on the plane.

Suppose , , has no zero. Its reciprocal is entire by the quotient derivative rule and U013, Corollary2.2. Since as , choose such that outside that disk. On the disk, continuity and nonvanishing give a positive minimum of . Thus is bounded on the whole plane and tends to zero at infinity. The preceding argument makes it identically zero, contradicting . Hence a root exists.

The finite telescoping identity $p(z)-p(r)=(z-r)\sum_{k=1}^d a_k \sum_{j=0}^{k-1}z^{k-1-j}r^j$ divides out that root and leaves a degree- polynomial with leading coefficient . Induction completes the factorization. At any root, the number of its occurrences in the product is exactly the first nonzero Taylor order, because all other factors are nonzero there. This proves the multiplicity statement and also the bound of on the number of distinct roots.

Compact distributions acting on Schwartz functions

For a multi-index , write $P_N(\phi)=\max_{|\alpha|\le N}\sup_{x\in\mathbb R^n} \langle x\rangle^N|\partial^\alpha\phi(x)|, \qquad \langle x\rangle=(1+|x|^2)^{1/2}. \tag{C1}$ These seminorms define the Schwartz topology by F1. Compact distributions act on smooth functions through a cutoff equal to one near their support, as proved in U021, B0. All pairings below are complex bilinear.

Proposition 0.2 (entire transforms and two test operators). Let . Its transform is the regular tempered distribution represented on real space by the entire function $H_v(z)=\langle v(y),e^{-iy\cdot z}\rangle,\qquad z\in\mathbb C^n. \tag{C2}$ There is an integer such that, for every multi-index , $|\partial^\alpha H_v(\xi)|\le C_\alpha\langle\xi\rangle^k, \qquad \xi\in\mathbb R^n. \tag{C3}$ Multiplication by and the operator $L_v\phi(x)=\langle v(y),\phi(x+y)\rangle \tag{C4}$ are continuous maps . For each , both output seminorms are bounded by a constant times .

Proof. Fix a smooth compact cutoff equal to one near , with support in . Continuity of on that fixed compact set gives $|\langle v,f\rangle| \le C\max_{|\beta|\le k}\sup_{|y|\le R} |\partial_y^\beta(\kappa f)(y)|. \tag{C5}$ The finite-order construction in the proof of U059, Theorem 3.1, establishes the entire function for every compact distribution : exponential Taylor series and all derivatives through its finite order converge uniformly on compact parameter sets, and the real restriction represents its inverse transform. That part of the proof uses only the finite-order estimate, independently of the later hypothesis. Apply it with , replace by , and multiply by . It gives (C2), including its real distributional interpretation and $\partial_z^\alpha H_v(z) =\langle v(y),(-iy)^\alpha e^{-iy\cdot z}\rangle. \tag{C6}$ In (C5), each derivative through order of the last integrand, including the cutoff, has at most factors from . The bounded -polynomials only change the constant depending on . This proves (C3) with the same for every . The product rule, applied through order , proves .

U021, B1, permits every -derivative to pass through the compact pairing in (C4). For , the elementary triangle bound gives . Apply (C5) to each , . Its needed derivatives have order at most , and its weight of degree is bounded by the input weight of degree . This proves . Both maps are therefore continuous.

Proposition 0.3 (a compact convolution factor). For and , their distributional convolution is tempered and $\langle u*v,\phi\rangle=\langle u,L_v\phi\rangle,\qquad F(u*v)=H_v\,Fu. \tag{C7}$ For fixed , the map is continuous for both the weak and strong dual topologies of . For every , one also has, as an identity in , $v*e^{ix\cdot z}=H_v(z)e^{ix\cdot z}. \tag{C8}$ Complex plane waves in (C8) need not be tempered.

Proof. Proposition 0.2 makes the right side of (C7) a continuous functional on . For a compactly supported , is supported in the compact set . U021, B3 and Theorem 1.1, then identifies this pairing with the proper-addition definition of : one compact factor makes the addition map proper on the two supports. Thus it is the original convolution on compact tests. F1's density of compact tests in makes this tempered extension unique.

The needed test identity is $F(H_v\phi)(x)=\langle v(y),F\phi(x+y)\rangle =L_v(F\phi)(x). \tag{C9}$ Indeed, insert (C2) in the integral on the left. Moving past that integral is justified by (C5): after a cutoff in , every -derivative through order of is bounded by , integrable in . On bounded integration sets the smooth-parameter Riemann-sum argument in U021, B1, applies. The displayed integrable majorant makes the tails tend to zero in every one of the finitely many seminorms needed by , proving the interchange over the whole space. Both sides of (C9) are Schwartz functions by Proposition 0.2 and F2. Consequently $\langle F(u*v),\phi\rangle =\langle u,L_vF\phi\rangle =\langle u,F(H_v\phi)\rangle =\langle H_vFu,\phi\rangle. \tag{C10}$ Here the transpose convention is exactly F3.

For weak continuity, each output test is the fixed input test . For strong continuity, let be any bounded subset of . Proposition 0.2 makes bounded and gives the exact seminorm identity $\sup_{\phi\in B}|\langle u*v,\phi\rangle| =\sup_{\psi\in L_vB}|\langle u,\psi\rangle|. \tag{C11}$ This proves continuity for the topology of uniform convergence on bounded sets.

Finally, U021, B1 and Theorem 1.1, identify convolution of a compact distribution with a smooth function as the smooth function obtained by the compact parameter pairing. Apply this to . Its factor is independent of , so the remaining pairing is . Compact tests suffice for this argument, whatever the imaginary part of .

Every compact distribution with only trivial convolution factors

Call a factor trivial when it is a nonzero scalar multiple of a shifted point mass. A factorization is trivial when at least one of its factors is trivial. These are the units of the compact-distribution convolution algebra: has inverse . Treating only the mass-one as trivial would make scalar refactorizations of every nonzero input fail the intended condition; the algebraic unit convention makes that distinction explicit.

Theorem 1.1. A nonzero has only trivial convolution factorizations , , if and only if $\begin{gathered} u=c\delta_a+d\delta_a',\\ a\in\mathbb R,\quad c,d\in\mathbb C,\\ (c,d)\ne(0,0). \end{gathered} \tag{1.1}$

Every zero-free compact transform is a point mass. For , its entire Fourier transform $W(z)=\langle w(x),e^{-izx}\rangle$ is justified by the same finite-order exponential-series construction as in Sharp bounds for compact spectra, formula (3.2), with the exponent sign changed and the inverse-transform normalization omitted. That argument uses only compact finite order for its entire construction and test interchange; the later norm estimates of that theorem are not required here. A compact cutoff supported in some and the finite-order estimate give $|W(z)|\le C(1+|z|)^N e^{A|\operatorname{Im}z|}. \tag{1.2}$ The real restriction is the whole tempered Fourier transform, by the absolute finite-order test interchange already proved in Sharp bounds for compact spectra, Theorem 3.1. If , Fourier injectivity gives .

Suppose has no complex zeros. The entire function has, by Cauchy kernels and distributional boundary limits's Cauchy coefficients, a Taylor series about zero that converges on every disk. Integrating that series termwise gives an entire primitive with . Choose a complex number with , using its nonzero polar form, and put . Differentiation shows that is constant and its value at zero is one, so on the whole plane.

Write . Formula (1.2) implies $\operatorname{Re}J(z) \le\log C+N\log(1+|z|)+A|z|.$ For each , denote the right side at radius by . On the circle , the real function is nonnegative. Uniform convergence of the Taylor series on this circle, obtained from any larger Cauchy circle, gives its mean . For , the -th positive Fourier coefficient of is . Consequently $\begin{gathered} \frac{|a_k|R^k}{2}\\ \le\frac1{2\pi}\int_0^{2\pi}q_R(\theta)d\theta\\ =M_R-\operatorname{Re}a_0. \end{gathered} \tag{1.3}$ For every , let ; the right side is , so . Therefore , with . On the real axis (1.2) is merely a polynomial bound. Applying it in both directions and forces . Write , . Fourier injectivity now gives . This proves the entire zero-free assertion, including the reality of the point location, without an unsupported entire-factorization theorem.

A complex zero gives an actual compact factor. Suppose has a zero at . Set , a compact distribution. Its total mass is . Compact forcing, moments and positive error kernels's complete compact-primitive theorem, with and order one, produces a compact distribution with , supported in any closed interval containing the support of . Put , still compact. The full distributional product rule gives $(D-z_0)v =-i e^{iz_0x}R' =u. \tag{1.4}$ Thus, with , $u=b_{z_0}*v.$ Compact-factor convolution and differentiation are the complete Convolution as addition of supports interfaces. The entire transform of is , so this factor is not a unit. The factor is nonzero because , and its entire transform is precisely the holomorphic quotient , by (1.4). Compactness was proved directly by the moment theorem, rather than asserted from a growth theorem.

If every factorization of is trivial, then either is zero-free, in which case the preceding result gives , or it has a complex zero. In the latter factorization must be a unit, since is not. Write and use (1.4): $u=-C_0z_0\delta_a-iC_0\delta_a'.$ This is (1.1). It proves the complete necessity.

Sufficiency, including every complex multiplicity. If has the form (1.1), its entire transform is $U(z)=(c+idz)e^{-iaz}. \tag{1.5}$ For any compact factorization , neither factor is zero. Their entire transforms multiply to : the compact convolution pairing applied to separates into the two finite-order pairings. If , (1.5) has no zeros, so both factor transforms are zero-free. If , it has exactly one zero, of multiplicity one. Both factor transforms cannot have a zero: different zero locations would give two distinct zeros of the product, while the same location would give multiplicity at least two. The multiplicities add because each local Taylor series factors into its first nonzero power times a nonvanishing analytic function. Hence at least one factor is zero-free. The proved zero-free result makes that compact factor a nonzero scalar shifted point mass. This proves sufficiency in every case. Finally, any unit has a zero-free entire transform because its product with its inverse transform is one; the proved zero-free result and the explicit inverses above show that the scalar shifted point masses are all the units.

When the linear factors give a finite factorization

Theorem 1.1 supplies a factor at each complex zero. A finite list of zeros has a particularly concrete meaning in physical space.

Theorem 1.2 (finite zeros and point jets). For nonzero , the following conditions are equivalent: has finitely many zeros counted with multiplicity; is a finite point jet at one real point; and is a finite convolution of units and the linear nonunit factors in Theorem 1.1. If its zeros are , repeated according to multiplicity, then $\begin{gathered} u=C\prod_{\ell=1}^m(D-z_\ell)\delta_a,\\ Fu(z)=C e^{-iaz}\prod_{\ell=1}^m(z-z_\ell),\\ C\ne0,\qquad a\in\mathbb R. \end{gathered} \tag{K1}$ The zero-free case has and is a unit.

Proof. At an isolated zero, the convergent Taylor series of a nonzero entire function has a first nonzero term of finite degree. The compact factor construction in the preceding proof removes one occurrence of that zero: it gives an actual compact with . The Taylor series shows that the multiplicity decreases by one there, and the quotient creates no other zeros. It is nonzero by this identity. Repeating through the finite list leaves a compact factor whose transform is zero-free. Theorem 1.1's proved zero-free classification makes it , giving (K1) with the original Fourier signs.

If the product polynomial is , the ordinary point derivatives in (K1) have coefficients , because . Conversely a nonzero finite jet has transform . At least one coefficient is nonzero because the given distribution is nonzero; thus that polynomial is nonzero. Lemma0.1 supplies its leading coefficient and every root with multiplicity, so it gives a finite convolution as in (K1). Finally each classified nonunit is a first-order point jet, and finite convolutions of point jets and units remain point jets at the sum of their real support points. This proves all implications.

The forcing , , has zeros for every integer . Each zero supplies a compact linear factor, but Theorem 1.2 excludes a finite factorization solely into the classified linear factors and units. The distinction concerns the complete zero set, including every multiplicity.

Homogeneous equations: compact solvability and smooth density

Theorem 2.1. Let be a homogeneous complex polynomial in variables, and let $\mathcal N_P=\{v\in C^\infty(\mathbb R^n):P(D)v=0\}.$ Then its polynomial members are dense in in the topology of uniform convergence of every derivative on every compact set. For , there exists with $P(D)q=f \tag{2.1}$ if and only if for every polynomial satisfying . Both assertions include the zero polynomial and the nonzero constant case.

We first prove the analytic mechanism and then turn the moment condition into exactly the required quotient.

A uniform division estimate on short complex lines

Let be homogeneous of degree . Choose a real vector with . Such a vector exists: a polynomial zero at every real point is zero, by induction on the number of variables, applying the one-variable finite root bound to each coefficient in the remaining variables. For every , the one-variable polynomial has degree and the fixed leading coefficient . Lemma0.1 writes it as $P(z+\tau a)=P(a)\prod_{\ell=1}^m(\tau-\rho_\ell),$ retaining roots with all multiplicities. Set . Remove from the intervals of distance less than from the real numbers . Their total length is at most , so there is outside them. On the entire circle , $\begin{gathered} |P(z+\tau a)|\ge c_P,\\ c_P=|P(a)|\delta^m>0. \end{gathered} \tag{2.2}$ The radius may depend on ; the lower bound does not.

If with a polynomial or an entire function, the one-variable Cauchy mean formula for gives $|Q(z)|\le c_P^{-1} \sup_{|\tau|=r}|F(z+\tau a)|. \tag{2.3}$ This is also valid at zeros of , because it applies to the already analytic , without evaluating a singular pointwise quotient there.

In particular, suppose satisfies the whole growth estimate $\begin{gathered} |F(z)|\le C(1+|z|)^N\\ {}\times e^{A|\operatorname{Im}z|},\qquad A\ge0. \end{gathered} \tag{2.4}$ If its quotient is entire, (2.3), and give $|Q(z)|\le C'(1+|z|)^N e^{A|\operatorname{Im}z|}. \tag{2.5}$ Only the constant changes: and . Thus division keeps polynomial growth on the real subspace and finite exponential growth in the imaginary directions.

A compact inverse transform from that growth

We prove the exact range statement needed here. Let be any entire function on satisfying (2.5). Its real restriction has polynomial growth, so it is tempered, and is a well-defined tempered distribution. In fact $\operatorname{supp}q\subset[-A,A]^n. \tag{2.6}$ This coordinate bound establishes compactness. The next lemma gives the sharper Euclidean radius with the same exponent.

For , put $q_\epsilon(x)=(2\pi)^{-n} \int_{\mathbb R^n}Q(\xi) e^{-\epsilon|\xi|^2}e^{ix\cdot\xi}d\xi.$ The integral is absolute, and identifies its function distribution with the inverse transform of , by absolute Fubini against a Schwartz test. As , the latter functions converge to in : the polynomial bound times any Schwartz test is integrable and dominates their differences. Fourier continuity therefore gives distributionally.

Fix a coordinate , a real vector , and a real height . Shift the -th integration line from to . The full identity is $\begin{gathered} \zeta=\xi+it e_j,\\ q_\epsilon(x)=(2\pi)^{-n}\int_{\mathbb R^n}Q(\zeta)\\ {}\times e^{-\epsilon\zeta\cdot\zeta+ix\cdot\zeta}d\xi. \end{gathered} \tag{2.7}$ Here is the contour justification, including the unbounded integration. Hold the other real coordinates fixed and use a rectangle with vertical sides at and the two horizontal heights . Its integrand is entire in that coordinate. Its closed integral is zero by the complex Green identity from Boundary flux and weak identities, Corollaries2.3–2.4: first round the four corners to apply the smooth-boundary identity, then let their radii tend to zero; continuity makes the four removed arc contributions tend to zero, and the area Cauchy–Riemann term is identically zero. On a vertical side, with height between , the modulus is bounded by a polynomial in times $\exp(-\epsilon R^2-\epsilon|\xi_{\widehat j}|^2 +\epsilon t^2+A|t|+|x_jt|).$ For fixed , its integral over and over the other coordinates tends to zero as ; the Gaussian dominates that polynomial. The horizontal tails also tend to zero by the same majorant. Fubini is absolute in all these bounded-height integrals. This proves (2.7) for each fixed finite , of either sign.

Formula (2.5) in (2.7) gives, with constants independent of , $\begin{gathered} |q_\epsilon(x)|\le C''\epsilon^{-(N+n)/2}\\ {}\times(1+|t|)^N e^{-x_jt+A|t|+\epsilon t^2}. \end{gathered} \tag{2.8}$ For the integral bound use , then the substitution ; the integral of is finite.

If , choose . If , choose . Both choices make the last exponential in (2.8) equal $\exp\!\left(-\frac{(|x_j|-A)^2}{4\epsilon}\right).$ On every compact subset of , its distance from the box boundary is positive and is bounded. Thus the polynomial factor in (2.8) is bounded by a fixed power of , while this exponential tends to zero faster than any such power. Consequently uniformly there. Distributional convergence now gives on that open coordinate region. The complement of is the finite union of these coordinate regions; local smooth partitions of unity, or the locality of a zero distribution, give (2.6). In particular is a compact distribution. This proves the full inverse-transform mechanism used below.

The unchanged radial exponent gives a Euclidean ball

The coordinate argument also gives the sharper geometric interpretation of the same constant .

Lemma 2.2 (radial support radius). Under (2.5), the inverse transform satisfies $\operatorname{supp}q\subset\overline B(0,A). \tag{K2}$

Proof. Fix a real unit vector , and choose a real orthogonal coordinate matrix whose first coordinate direction is . To construct this matrix, extend the nonzero vector to a real basis and apply the proved Gram process in the supplied finite algebra, §§10.1 and10.5, beginning with ; its first normalized vector stays . The resulting matrix satisfies , so determinant multiplicativity gives . Its real Jacobian therefore has absolute value one; it preserves the Euclidean norm and, after complex linear extension, the bilinear quadratic form . In these coordinates the complete rectangle proof of (2.7) shifts the first line by , or equivalently shifts the original frequency vector to , for each fixed .

The vertical sides still have Gaussian decay in their real coordinate and in all remaining real coordinates, multiplied by a polynomial at fixed finite height. Thus their integrals vanish, the horizontal tails converge, and all integrations are absolute exactly as in that proof. Applying the radial bound gives $\begin{gathered} |q_\epsilon(x)|\le C_v\epsilon^{-(N+n)/2}(1+t)^N\\ {}\times e^{-t(v\cdot x-A)+\epsilon t^2},\\ 0<\epsilon\le1,\qquad t\ge0. \end{gathered} \tag{K3}$ No uniformity of the constant in is needed. On a compact subset of , take . The last exponential is , with a uniform positive gap. It dominates every remaining inverse power of , so uniformly there. The already proved distributional convergence gives on this open half-space.

For any with , take . The strict half-space contains a neighborhood of . Thus that point is outside the support of , proving (K2), including . The exponent was never enlarged.

The moment condition produces an entire quotient

Let annihilate every polynomial in . Its entire Fourier transform $F(z)=\langle f(x),e^{-iz\cdot x}\rangle =\sum_{k\ge0}F_k(z)$ has the homogeneous Taylor pieces $F_k(z)=(-i)^k \sum_{|\alpha|=k} \frac{\langle f,x^\alpha\rangle}{\alpha!}z^\alpha. \tag{2.9}$ The compact finite-order estimate proves entire convergence and (2.4), by the entire compact-Fourier construction in Sharp bounds for compact spectra, Theorem 3.1, and the preceding proof: choose a compact cutoff supported in a ball of radius ; derivatives of the exponential give the polynomial factor, and its modulus is at most . Choose a larger if necessary. The real restriction is exactly .

Let be the finite-dimensional space of homogeneous polynomials of degree . The bilinear coefficient pairing $B_k(R,p)=p(\partial_z)R(0) =\sum_{|\alpha|=k}\alpha!r_\alpha p_\alpha$ is nondegenerate. For , (2.9) says . For , kills all of , so . For , put $A_k=P(D):\mathcal H_k\longrightarrow\mathcal H_{k-m}.$ The finite derivative identity, with all factors retained, is $\begin{gathered} B_{k-m}(R,A_kp)\\ =(-i)^m B_k(PR,p). \end{gathered} \tag{2.10}$ It follows by expanding , and into their monomials: the factorial remaining after differentiation is exactly the coefficient pairing on the right.

The functional vanishes on by the moment hypothesis. It therefore factors through . Extend that finite-dimensional linear functional to by basis extension, and represent it as , using nondegeneracy. Formula (2.10) then gives $\begin{gathered} F_k=P Q_{k-m},\\ Q_{k-m}=(-i)^mR\in\mathcal H_{k-m}. \end{gathered} \tag{2.11}$ The quotient polynomial is unique because the complex polynomial ring has no zero divisors: the product of the leading monomials in any fixed lexicographic order has a nonzero coefficient. The scalar factors in (2.9)–(2.11) are never suppressed.

It remains to prove that the formal quotient actually converges everywhere. Apply (2.3) to the polynomial identity (2.11). For , put . On the selected circle, . For every , the one-variable Cauchy formula applied to , , gives $\begin{gathered} \sup_{|\zeta|\le B}|F_k(\zeta)|\\ \le M_F(L)(B/L)^k,\\ M_F(L)=\max_{|\zeta|\le L}|F(\zeta)|. \end{gathered}$ Indeed the coefficient of is , its Cauchy bound is , and homogeneity supplies the factor . Thus $\begin{gathered} \sup_{|z|\le R_0}|Q_{k-m}(z)|\\ \le c_P^{-1}M_F(L)(B/L)^k. \end{gathered} \tag{2.12}$ The right side is a summable geometric sequence. Therefore converges uniformly on every complex compact ball. It is entire: use the same uniform bounds on a slightly larger ball and apply the one-variable Cauchy derivative estimate successively in the coordinates to each polynomial term; every differentiated series is then locally uniformly convergent. Their sums satisfy the coordinate Cauchy–Riemann equations and the corresponding local multivariable power-series expansion. Multiplication by the fixed polynomial passes through the locally uniform series, giving everywhere.

Now (2.3) applies to that entire quotient and (2.4) gives (2.5). The proved inverse-transform statement supplies a compact . On real frequencies, ; the full distributional Fourier differentiation rule and injectivity give . This proves sufficiency in (2.1) for every nonzero homogeneous polynomial of positive degree.

Necessity retains the bilinear transpose sign. If with compact, then for every polynomial , $\begin{gathered} \langle f,p\rangle=\langle q,P(-D)p\rangle\\ =(-1)^m\langle q,P(D)p\rangle=0. \end{gathered} \tag{2.13}$ Compact support permits pairing with that global polynomial by a cutoff, and the distributional product and transpose rules give the displayed identity. The homogeneity is what makes . This proves both directions of the exact compact-solvability criterion.

Density in the full smooth topology

Let be the closure in of the vector space of polynomial null solutions. Continuity of makes closed, so . Suppose . The complete compact-distribution separation lemma Fundamental solutions, continuation and approximation, Lemma 3.1 produces with and . In particular it annihilates all null polynomials. The just-proved criterion gives a compact with . But (2.13), now applied to the smooth global null solution , gives $1=\langle f,v\rangle =(-1)^m\langle q,P(D)v\rangle=0,$ a contradiction. Therefore .

This is sequential approximation in exactly the stated topology. For each integer , choose a null polynomial so that $\max_{|\alpha|\le j}\sup_{|x|\le j} |\partial^\alpha(v-p_j)(x)|<1/j.$ Closure in the smooth seminorm topology permits this finite-seminorm choice. Every fixed compact set and derivative order is contained in these controls for all sufficiently large , so with every derivative on every compact set. No bound on the growth of , ellipticity, real coefficients or simplicity of the factors of was assumed.

If is a nonzero constant, its smooth null space and polynomial null space are both zero, the moment condition is empty, and is compact for every . If , all polynomials and all smooth functions are null. A compact annihilating every polynomial is zero by Compact forcing, moments and positive error kernels, Lemma 1.2, so the compact equation has a solution exactly under the stated condition. Polynomials are dense in the full smooth space: otherwise Fundamental solutions, continuation and approximation separation would produce a nonzero compact distribution annihilating them, contradicting that same complete moment-detection lemma. The identical seminorm selection gives sequential approximation in this case too. All degenerate cases are thus included.

Compact uniqueness and the actual support hull

Existence in Theorem 2.1 uses homogeneity. Once a compact solution exists, its uniqueness and support geometry hold for every nonzero polynomial.

Corollary 2.3. Let be any nonzero complex polynomial on . A compact solution of is unique. For nonzero forcing, $\operatorname{ch}\operatorname{supp}q =\operatorname{ch}\operatorname{supp}f, \tag{K4}$ where denotes the closed convex hull. For zero forcing the unique compact solution is zero.

Proof. Set . Its support is contained in , and its entire transform is the nonzero polynomial , so and its support is exactly . The differentiation and compact convolution rules give . The full compact convolution support theorem in Convex supports and convolution cancellation, Theorem 3.1, applies to arbitrary complex distributions and gives $\begin{gathered} \operatorname{ch}\operatorname{supp}(P(D)q)\\ =\{0\}+\operatorname{ch}\operatorname{supp}q. \end{gathered} \tag{K5}$ If , its hull is nonempty and the right side is nonempty, so . Apply this to the difference of two compact solutions to get uniqueness, and to a solution of nonzero forcing to get (K4). Nonzero constants are included.

Equality of hulls allows interior support to change. The compact primitive of has interval support, while the forcing has only two support points. This corollary does not extend the polynomial-moment existence criterion to nonhomogeneous symbols.

Approximation that preserves finitely many observations

Smooth convergence controls derivative values at finitely many points. Those values can also be kept exactly at every approximating stage, provided they belong to the null solution being approximated.

Corollary 2.4 (compatible finite constraints). Keep the homogeneous or degenerate scope of Theorem 2.1. For and any continuous complex-linear map , there are polynomial null solutions with $p_j\longrightarrow v\text{ in }C^\infty,\qquad Jp_j=Jv \tag{K6}$ for every . This includes any finite set of derivative values at finitely many points.

Proof. Let be the polynomial null space. Theorem 2.1 gives converging to . Choose a basis of . Finite elimination on the independent columns supplies coordinate rows such that the matrix is invertible. For , its basis coordinates are , a continuous function of . If converges to in , pass to the limit in to get that same identity for . Hence is closed.

Continuity gives , so . Choose with , and set $\begin{gathered} b_j=Jv-Js_j,\\ \lambda_j=A^{-1}(b_j)_I,\\ p_j=s_j+\sum_{\ell=1}^d(\lambda_j)_\ell h_\ell. \end{gathered} \tag{K7}$ Then is null and . Since , every smooth compact seminorm of the correction is bounded by . This proves full smooth convergence. If , the map vanishes on the closure and the original already has the required observations. The nonzero constant and zero-symbol cases follow from the corresponding cases of Theorem 2.1.

The target data are . A differential equation can make arbitrary jet assignments incompatible; the corollary preserves data of an actual solution.

Linear coordinates and homogeneous distributions

For an invertible real matrix , let and define $\langle A^*u,\phi\rangle =d^{-1}\langle u,\phi\circ A^{-1}\rangle. \tag{H3}$ Thus for a regular distribution. Write , defined on tests by translation by , and , where .

Proposition 3.1 (coordinate, modulation and dilation laws). These operations preserve , and $\begin{aligned} F(A^*u)&=d^{-1}(A^{-T})^*Fu,\\ F(\tau_hu)&=e^{-ih\cdot\xi}Fu,& F(M_hu)&=\tau_hFu,\\ F(D_ju)&=\xi_jFu,&F(x_ju)&=-D_{\xi_j}Fu . \end{aligned} \tag{H4}$ For , put . If for all , with and , then is homogeneous of degree .

Proof. The chain rule and , prove continuity of composition by either or on every Schwartz seminorm, allowing an increase in its index. Translations have the same property by the triangle inequality, and multiplication by has it by the product rule. Hence their transposes define continuous operations on .

An absolutely convergent change of variables on a Schwartz test gives $F(\phi\circ A^T)(x)=d^{-1}F\phi(A^{-1}x). \tag{H5}$ The left side of the first identity in (H4), paired with , is . The right side has the same pairing: cancels its displayed . This proves the absolute determinant and inverse transpose for every distribution, including singular ones.

The two test identities $\begin{aligned} F(e^{-ih\cdot\xi}\phi)(x)&=F\phi(x+h),\\ F(\phi(\,\cdot+h))(x)&=e^{ih\cdot x}F\phi(x) \end{aligned} \tag{H6}$ follow respectively by combining exponentials and substituting in the absolute integral. Transposition proves the translation and modulation laws. Integration by parts and differentiation of that same Schwartz integral, justified in F2–F5, give the last two laws with .

Set in the first law. Homogeneity gives , so . Replacing by gives degree , with no restriction on the complex number .

Theorem 3.2 (homogeneity away from the origin). Let , , and suppose its restriction to is homogeneous of degree . Then has a unique tempered extension of its action to Schwartz tests. If that restriction is represented by a smooth function, is also represented by a smooth function off the origin. In particular, Fourier transformation is a bijection between globally homogeneous distributions of degree that are smooth off zero and those of degree with the same smoothness property.

Proof. Use a smooth compact cutoff equal to one on and zero on , as constructed by the cutoff prerequisite. Put ; its support lies in the fixed annulus . Telescoping gives $1=\chi(x)+\sum_{j=0}^{\infty}\psi(2^{-j}x). \tag{H7}$ On each compact set the sum has only finitely many nonzero terms. The annular homogeneity definition in Homogeneous extensions and angular moments, Lemma 1.1, gives for $\langle u,\psi(\,\cdot/t)\phi\rangle =t^{a+n}\langle u,\psi\,\phi(t\,\cdot)\rangle \tag{H8}$ when is a compact test. This uses homogeneity only on tests supported away from zero.

Let be a finite order bound for on the fixed annulus. The product and chain rules imply, for , $\begin{aligned} |\langle u,\psi\,\phi(t\,\cdot)\rangle| &\le C\sum_{|\beta|\le k}t^{|\beta|} \sup_{y\in tK}|\partial^\beta\phi(y)|\\ &\le C_N P_N(\phi)t^{k-N},\qquad N\ge k . \end{aligned} \tag{H9}$ The first inequality includes the bounded derivatives of the fixed . The second uses on . Therefore the absolute value of the -th term in (H8) is at most $C_NP_N(\phi)\,2^{j(\operatorname{Re}a+n+k-N)}. \tag{H10}$ The compact distribution is bounded by another finite Schwartz seminorm, directly from its compact finite order. Choose an integer large enough for that bound and for and . Define the action on a Schwartz test by plus the series of the right sides of (H8). Estimate (H10) is a convergent geometric bound by , so this defines a tempered distribution. On compact tests (H7) proves agreement with the original . Compact-test density from F1 proves uniqueness. In particular, no condition on point-supported terms at the origin was needed.

Now suppose off zero, with smooth. The distributional homogeneity identity implies pointwise there. Indeed the difference is continuous and has zero integral against every compact test; multiplying a sufficiently small nonnegative bump by a constant complex phase detects any nonzero value. Differentiating the pointwise identity shows $|\partial^\alpha v(x)|\le C_\alpha|x|^{\operatorname{Re}a-|\alpha|}, \qquad |x|\ge1, \tag{H11}$ by boundedness of every derivative on the unit sphere.

Fix a nonnegative integer and a coordinate . Choose an integer . Decompose the tempered distribution into $\partial_j^m u=\chi\,\partial_j^m u+g,\qquad g(x)=(1-\chi(x))\,\partial_j^m v(x). \tag{H12}$ The first term is compact. The second is a smooth function vanishing near zero; by (H11), for . To verify the tail integral without a polar-coordinate assumption, partition into dyadic annuli: the annulus of radius has volume at most , whereas the integrand is at most . The exponents sum to a strictly negative number by the choice of . Thus the moments are integrable. The equality in (H12) holds first on compact tests and hence on Schwartz tests by density, since all terms are tempered.

The Fourier transform of the compact term is by Proposition 0.2. The ordinary Fourier transform of is : each derivative through order is the integral of ; the integrable moments dominate both continuity and the fundamental-theorem difference quotients. Absolute integration against a Schwartz test identifies this ordinary transform with the distributional one. Using Proposition 3.1, $(i\xi_j)^m Fu=F(\partial_j^m u)\in C^r(\mathbb R^n). \tag{H13}$ On , division by the smooth nonvanishing gives a representative of . As increases these representatives coincide: any two are continuous representatives of the same distribution, so the bump argument above forces equality. Since the coordinate sets cover the punctured space, is smooth there.

For global homogeneity, Proposition 3.1 gives the stated complementary degree. Applying the result again to degree , and using from F3, proves surjectivity as well as injectivity. Reflection preserves both degree and punctured smoothness, so is the inverse on these spaces.

Exercises

Exercise 1 (foundation: two opposite point sources). For , give an explicit nontrivial compact convolution factorization of through . Compute the entire quotient transform, including its removable value at zero.

Exercise 2 (intermediate: a second difference has a compact second primitive). For , set . Compute a compact with , its full transform and its support. Give a convolution factorization and retain every root multiplicity.

Exercise 3 (intermediate: two complex point-jet factors). For , , expand into ordinary point derivatives. Determine whether its compact convolution factorizations are all trivial, including .

Exercise 4 (foundation: an odd-order bilinear obstruction). On , set . For real points , determine exactly when has a compact distribution solution. Exhibit the detecting null polynomial and the transpose sign.

Exercise 5 (intermediate: zero mass and dipole moments are insufficient). On , let and . Show that its constant and linear moments vanish, yet has no compact solution. Compare with the forcing .

Exercise 6 (advanced: no finite degree cutoff for the moment test). For every integer , construct a compact distribution on annihilating every polynomial of degree at most , but having no compact solution under . Give one explicit higher-degree harmonic polynomial that detects it.

Exercise 7 (intermediate: smooth null functions can be nonanalytic). Determine every global smooth solution of on . Construct a sequence of polynomial null solutions converging with every derivative on every compact set, and explain why the construction covers a smooth function with a zero Taylor series at the origin that is not zero nearby.

Exercise 8 (advanced: all degenerate symbols and the role of homogeneity). State the compact-solvability criterion for the zero symbol and for a nonzero constant symbol. Then use to show that both the polynomial density and the compact-solvability criterion can fail if the homogeneity assumption is removed.

Exercise 9 (intermediate: uniqueness, hulls and a sharp radius). Prove conditional compact uniqueness and equality of support hulls for , despite the failure of the homogeneous existence criterion. Compare the supports of and its compact -primitive. Finally, show that for the radial exponent cannot be improved even by allowing a polynomial prefactor.

Exercise 10 (advanced: exact observations on a nonanalytic null solution). Put , where are arbitrary smooth functions. Given finitely many points and finitely many derivative observations, construct polynomial solutions of converging to in the full smooth topology while preserving those observations exactly. Include a smooth flat nonanalytic , and explain why a prescribed nonzero mixed derivative at a point is incompatible.

Exercise 11 (intermediate: a repeated complex factor is a point jet). For , and distinct , expand into ordinary point derivatives. Determine its full complex zero set and every multiplicity. Explain why , , cannot be a finite convolution of these classified first-order factors and units.

Exercise 12 (intermediate: complex plane waves for a compact difference). Let , , and on the line. Find every such that . Determine which of these plane waves are tempered and when real zeros exist.

Exercise 13 (intermediate: a shear in physical and frequency coordinates). For $A=\begin{pmatrix}2&1\\0&1\end{pmatrix},\qquad b=(1,-1),$ compute and the transform of . Explain separately the determinant factor and the covector location.

Exercise 14 (advanced: the two punctured hypotheses). On , use to distinguish homogeneity away from zero from global homogeneity, and compute . On , use to show that punctured homogeneity alone does not make the Fourier transform smooth away from zero.

Solutions

Solution 1. The ordinary derivative of is . Thus $\begin{gathered} q=i\boldsymbol1_{(0,a)},\\ Dq=\delta_0-\delta_a,\\ \delta_0-\delta_a=(D\delta_0)*q. \end{gathered}$ Both factors are compact and neither is a scalar shifted point mass: one is a derivative point jet and the other is a nonzero interval function. The entire input transform is , and $Fq(z)=\frac{1-e^{-iaz}}z, \qquad Fq(0)=ia.$ The finite interval integral gives this quotient and its entire extension directly. In particular the zero at the origin has been divided out with the correct Fourier and factors.

Solution 2. Put as above. Then , where $\ell_a(x)= \begin{cases} x,&0\le x\le a,\\ 2a-x,&a\le x\le2a,\\ 0,&x\notin[0,2a]. \end{cases}$ Its slope jumps are , so . Since , . The support is exactly , and $Fq(z)=\left(\frac{1-e^{-iaz}}z\right)^2, \qquad Fq(0)=-a^2.$ Also , with two nonunit compact factors. Its transform has double zeros at , every ; the numerator's first derivative is nonzero at each such point before squaring. The quotient removes exactly the double zero at zero and is entire.

Solution 3. Using , the expansion is $u=-\delta_a''+i(z_1+z_2)\delta_a' +z_1z_2\delta_a.$ It factors as . Each entire factor transform has a zero, so neither factor is a unit by Theorem 1.1's complete classification. Their product is . When the two points are distinct it has two simple zeros; when they coincide it has one double zero. In both cases the factorization is nontrivial. The second derivative coefficient is , so the distribution is nonzero even in the coincident case.

Solution 4. The polynomial satisfies , , and . Its pairing with the forcing is , nonzero exactly when because the coordinates are real. If a compact existed, the bilinear transpose would give $\begin{gathered} \langle P(D)q,p\rangle\\ =\langle q,P(-D)p\rangle\\ =-\langle q,P(D)p\rangle=0. \end{gathered}$ This excludes every distinct pair. If , the forcing is zero and is compact. This proves the full characterization; no conjugation has entered the polynomial or the transpose.

Solution 5. The total mass is . The two first coordinates cancel and all second coordinates are zero, so both linear moments vanish. But is a null polynomial for , and . Theorem 2.1's necessary compact-moment condition rules out the solution. In contrast has the compact solution . For every harmonic polynomial , its pairing is , with positive transpose sign because the degree is two. Thus the entire moment condition, rather than just its first two degrees, distinguishes these forcings.

Solution 6. Set and . Every polynomial of degree at most has zero -th derivative, so all the indicated moments vanish. The real polynomial $p(x_1,x_2)=\operatorname{Re}(x_1+ix_2)^m$ is harmonic: differentiating the complex polynomial twice in the two coordinates gives opposite terms, including , where both second derivatives are zero. Its coefficient of is one, hence . This null polynomial detects the obstruction in Theorem 2.1, so no compact Laplace-type solution exists. The argument works for every finite ; a finite truncation of the polynomial-moment condition cannot replace all degrees.

Solution 7. The equation is . Twice applying the real fundamental theorem, including oriented integrals when a coordinate is negative, gives $u(x_1,x_2)=u(x_1,0)+u(0,x_2)-u(0,0).$ Thus every solution is , and every such sum solves the equation. The zero-symbol density case of Theorem 2.1 in one dimension supplies polynomial sequences approximating these smooth functions with all derivatives on compact intervals. Choose their -th errors below through order on , and set . Each mixed derivative is zero, while the pure derivative errors are controlled by the one-dimensional errors. These are polynomial null solutions converging in the full two-dimensional smooth topology.

For example take for , and zero for , with . Repeated differentiation on the positive side gives a polynomial in times that exponential, tending to zero at in every order; the negative side is zero. Thus this is smooth with zero Taylor series there but is positive arbitrarily nearby. The approximation uses smooth density, not convergence of its own Taylor series.

Solution 8. For , every polynomial is null. The compact moment-detection theorem makes the condition equivalent to , precisely the condition for the compact equation ; any compact , including zero, then solves it. For a nonzero constant , only the zero polynomial is null, the condition is empty, and is compact for every compact .

For the nonhomogeneous example , a nonzero polynomial cannot be null: its highest-degree part survives the constant term and cannot be cancelled by the Laplacian, which lowers degree by two. Yet the smooth function is null and is nonzero at the origin. The zero polynomial space cannot approximate it even pointwise there. Finally the forcing annihilates every polynomial null solution, since there are none except zero. A compact solution would have an entire transform satisfying $(z_1^2+z_2^2+1)Q(z)=1.$ The real-frequency identity extends to all complex coordinates by the identity theorem. Evaluating at gives , a contradiction. Thus both extensions fail when homogeneity is omitted, even though all finite-order transpose operations remain well defined.

Solution 9. The symbol is nonzero, so Corollary 2.3 gives uniqueness whenever a compact solution exists and equality of the actual hulls for nonzero forcing. This applies to this nonhomogeneous polynomial without an existence assertion: for example has solution , while Solution 8 proves that has no compact solution. These two conclusions are compatible. For the one-dimensional forcing, Solution 1 gives ; its support is , whereas the forcing support is . Both hulls are .

For a real point , , so its modulus is . If , evaluate a hypothetical bound with at : $\begin{gathered} e^{t(|b|-A)}\le C(1+t)^N,\\ t>0. \end{gathered} \tag{K8}$ Taking logarithms and dividing by contradicts , because . Thus is sharp. If , the admissible nonnegative radius is already zero.

Solution 10. Every such satisfies the equation because its mixed derivative is zero. Let collect the given finite derivative observations. They are continuous complex-linear functionals in the smooth compact topology. Theorem 2.1 supplies polynomial null . Form , choose its basis, invertible coordinate minor and null-polynomial lifts as in Corollary 2.4, and use exactly (K7). This is an explicit finite correction after each approximation: it preserves all observations, remains a polynomial null solution, and tends to zero in each smooth seminorm. If the observation image is zero, no correction is needed.

Take for , zero otherwise, and . Every positive-side derivative is a polynomial in times , tending to zero at the origin; this verifies smoothness and flatness in every order, while is positive arbitrarily close on the right. The construction therefore includes a nonanalytic solution and does not use its Taylor series as the approximating sequence. Conversely any null solution has at every point, so prescribing that mixed derivative to be nonzero is incompatible with the equation. Corollary 2.4 prescribes the actual data of .

Solution 11. Expand the polynomial before inserting the differential signs: $\begin{gathered} (z-r)^2(z-s)\\ =z^3-(2r+s)z^2\\ {}+(r^2+2rs)z-r^2s,\\ u=C\bigl[i\delta_a^{(3)}+(2r+s)\delta_a''\\ {}-i(r^2+2rs)\delta_a'-r^2s\delta_a\bigr]. \end{gathered} \tag{K9}$ Here , , and ; thus every ordinary derivative coefficient has its correct sign. The highest coefficient is nonzero. The transform is , with exactly a double zero at and a simple zero at , since the exponential never vanishes and . Its compact convolution has two copies of , one , and the unit . In contrast has a simple zero at every . There are infinitely many of them, so Theorem 1.2 excludes a finite convolution solely of classified first-order factors and units for that forcing.

Solution 12. Direct compact pairing gives . Write . Its zero condition separates into and . Thus precisely $z=\frac{\theta+2\pi k}{h}-\frac{i\log\rho}{h}, \qquad k\in\mathbb Z, \tag{C12}$ work, by (C8). A different choice of only reindexes the integers. Every zero is simple because there. If , all zeros are real and their bounded plane waves are tempered. If , every zero has nonzero imaginary part, so its wave grows exponentially along one direction and fails to be tempered. For a direct test, choose with ; such a bump exists by the continuous nonzero integrand. Pair with . Its absolute pairing is a fixed positive constant times , whereas each fixed Schwartz seminorm of the translated bump grows at most polynomially in . Taking toward the exponentially growing direction contradicts every possible finite-seminorm bound. Hence real zeros and tempered eigenwaves occur exactly when .

Solution 13. Here , so (H3) gives , whose Fourier transform is the constant . For the plane wave, , so its transform is . One can also apply (H4) to : the pullback by sends to , which cancels the outside factor . To verify that pullback directly, pair it with in (H3) using ; the result is . A frequency delta therefore changes its location by , while a regular transformed function is composed with . These are the same distributional identity.

Solution 14. Off zero, is the constant function and has degree zero. Globally, , so it is not homogeneous: the nonzero constant restriction would force degree zero, contradicted by the delta coefficient when . The transform is $Fu=(2\pi)^n\delta_0+1. \tag{H14}$ It is smooth off zero, as Theorem 3.2 requires, but has degree zero there rather than degree . This demonstrates why the degree conclusion requires global homogeneity.

For , substitution in its dilation pairing gives . It is tempered by a weighted supremum bound on that integral. Its Fourier transform is $Fw=2\pi\delta_0(\xi_1)\otimes1(\xi_2). \tag{H15}$ For completeness, is the one-dimensional transform in of the Schwartz function . The Schwartz property follows by integrating a weight of degree two more than the desired weight. One-dimensional inversion at zero gives , which is (H15) on every two-dimensional Schwartz test. Both and are singular along a whole line. For example, near a point with , a continuous representative of would vanish off that line, hence everywhere by continuity, but a compact test meeting the line has nonzero pairing. Thus is not smooth on the punctured plane; does not satisfy the punctured-smoothness hypothesis.

References

Convolution as addition of supports, B0–B3 and Theorem1.1: compact actions, locality and convolution. Compact forcing, moments and positive error kernels, Lemma1.2 and Theorem1.3: complete moment detection and compact primitives. Fundamental solutions, continuation and approximation, B1–B2 and Lemma3.1: smooth topology, compact duality and separation.
Cauchy kernels and distributional boundary limits, Theorem1.1 and Corollary2.2: Cauchy and Taylor formulas. Boundary flux and weak identities, Corollaries2.3–2.4: complex Green, including corners. Sharp bounds for compact spectra, Lemma0.1 and the finite-order construction in Theorem3.1: identity principle and entire compact transforms. Convex supports and convolution cancellation, Theorem3.1: exact convex hulls for arbitrary complex compact distributions.
Supplied finite algebra, §§10.1–10.5; Fourier foundation, F1–F5; functional foundation, §5; and the scalar and integration foundations named above. These exact copies retain their stated licences. Lemma0.1 supplies the additional complete polynomial factorization proof.
Lars Hörmander, The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis, reprint of the second edition (1990), Exercises7.3.1 and7.3.5, printed page390, with answers on pages414–415. The compact factor construction, entire quotient argument, exact support radius, finite-observation correction and fourteen graded problems above are independently expressed.
Homogeneous extensions and angular moments, Lemma 1.1: positive dilations on punctured tests. Lars Hörmander, The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis, reprint of the second edition (1990), Theorems 7.1.14–7.1.16 and 7.1.18, printed pages 166–168: compact transforms, convolution, Fourier coordinate laws and homogeneous distributions.