Choosing polynomial and exponential approximants

Written by GPT-6.1 Sol (OpenAI), Ultra reasoning effort, October 2026. Self-checked by GPT-6.1 Sol (OpenAI). Public domain (CC0).

Exponential-polynomial solutions supply a dense family on an open convex set. Sometimes their polynomial factors can be discarded. Sometimes their exponential factors can be discarded. These simplifications depend on different algebraic properties of the symbol. Multiplicities determine whether plane exponentials suffice; the location of the irreducible factors at the origin determines whether polynomials suffice.

This lesson uses Approximation and global solvability from support geometry, together with the compact-support Fourier inputs specified below. The original approximation theorem and polynomial criterion appear in Malgrange's paper [Malgrange]. We prove the consequences in arbitrary intersections of the weighted spaces used in this course. The intermediate complex analysis also explains why a globally irreducible polynomial may have several local analytic branches.

Fix a nonzero polynomial $P\in\mathbb C[z_1,\ldots,z_n]$, use $D=-i\partial$, and retain complex-linear distribution pairings. Let $X\subset\mathbb R^n$ be nonempty, open, and convex. Put $\begin{gathered} \mathcal F(X)=\bigcap_{i\in I}B_{p_i,k_i}^{\mathrm{loc}}(X),\\ I\ne\varnothing,\qquad 1\le p_i<\infty. \end{gathered} \tag{1}$ with moderate weights and the topology generated by the cutoff norms. Write $\mathcal N_P(X)$ for its closed subspace of solutions of $P(D)u=0$.

Primitive factors and moving polynomial roots proves unique factorization, Gauss’s lemma, rational-function Bézout and the holomorphic simple-root charts. Cauchy bounds, root counts and analytic extensions supplies Cauchy estimates, compact-contour differentiation, maximum modulus and persistent disk root counts. The separation argument is Continuous functionals, test families and compact limits, Section 1.

Two algebraic criteria

Factor the nonconstant part of the symbol over $\mathbb C$: $P=c\prod_{j=1}^{s}r_j^{m_j}, \qquad c\ne0, \tag{2}$ where the $r_j$ are distinct nonconstant irreducible polynomials and $m_j\ge1$. Constant factors are units. A nonzero constant $P$ has the empty factorization and $\mathcal N_P(X)=\{0\}$.

Theorem 1.1. In the topology inherited from (1), the following statements hold.

Polynomial solutions are dense in $\mathcal N_P(X)$ if and only if $r_j(0)=0$ for every $j$.
Finite linear combinations of the plane exponentials $x\longmapsto e^{ix\cdot z},\qquad P(z)=0, \tag{3}$ are dense in $\mathcal N_P(X)$ if and only if every $m_j=1$.

The empty factorization satisfies both conditions. The necessity assertions hold on every nonempty open $X$, without convexity. Convexity enters the sufficiency proofs through the support of a compact inverse.

For $n=1$, the criteria can already be seen from ordinary differential equations. The kernel of $D^2$ is spanned by $1,x$; plane exponentials give only $1$. The kernel of $D(D-1)$ is spanned by $1,e^{ix}$; its polynomial solutions give only $1$. The multidimensional proof must also account for analytic continuation along the zero set of each irreducible factor.

The Fourier inputs and continuous annihilators

For compact $\mu$, set $F(\zeta)=\widehat\mu(\zeta) =\mu\bigl(e^{-ix\cdot\zeta}\bigr). \tag{4}$ This is entire. The transpose of $P(D)$ is $P(-D)$, so the divisor relevant to (4) is $P(-\zeta)$.

We use the following precise prerequisites from compact-support Fourier analysis.

$\mu$ annihilates every polynomial solution of $P(D)h=0$ if and only if $F(\zeta)/P(-\zeta)$ is holomorphic in a neighborhood of $0$.
$\mu$ annihilates every exponential-polynomial solution $e^{ix\cdot z}A(x)$ of the equation if and only if $F(\zeta)/P(-\zeta)$ is entire.
The quotient is entire if and only if there is a compactly supported distribution $v$ with $P(-D)v=\mu$. This $v$ is unique, and $\operatorname{ch}\operatorname{supp}v =\operatorname{ch}\operatorname{supp}\mu. \tag{5}$

The entire exponential-polynomial annihilator equivalence and compact Fourier division are proved in Compact Fourier division and multiplicity-sensitive annihilators, CF5.1 and CF2.1–CF4.1, under its exact declared Fourier and Cauchy inputs. The local polynomial-annihilator equivalence near zero and convergence of its formal quotient are proved in Polynomial tests and a convergent local Fourier quotient, LP0–LP7, relative to its declared compact-transform, finite polynomial and Cauchy entries. The lower scope of Distributions, kernels and analytic singularities remains explicit. Entire divisibility has not established convergence of an analytic germ from merely formal local divisibility. The support-hull equality is Convex supports and convolution cancellation, Corollary 4.3, after the D convention is substituted into its polynomial. CF4.1 supplies an additional proof of the ordinary support-hull statement; the compact singular-support hull remains separate. The algebraic continuation and weighted density arguments proved here use these exact inputs. Malgrange [Malgrange] is the classical reference for the approximation setting.

Let $L$ be a continuous functional on $\mathcal F(X)$. Lemma 2.1 of the preceding lesson gives a smooth restriction $\mu\in\mathcal E'(X)$. If the entire quotient criterion supplies $v$, (5) places its support in $X$. The same lemma then gives $L(u)=0\qquad\text{for every }u\in\mathcal N_P(X). \tag{6}$ Its cutoff and mollification argument is essential: a nonsmooth $u$ is not paired directly with $\mu$. Consequently, to prove either density assertion it suffices to show that the corresponding annihilation assumption makes the quotient in (4) entire. Hahn–Banach separation then applies to the closed solution subspace, or equivalently extends its continuous separating functional to $\mathcal F(X)$.

Extending a quotient from generic complex lines

We first prove the analytic facts used in both sufficiency arguments. Polynomial unique factorization, Gauss's lemma over the rational-function field, the Euclidean algorithm for one-variable polynomials, and the Cauchy integral formula are the entry algebra and complex analysis.

After an invertible linear change of complex coordinates $z=(t,w)$, every factor $r_j$ in (2) is monic as a polynomial in $t$, after division by a nonzero constant. To choose this change, select a real direction $\theta$ on which none of the finitely many highest homogeneous parts of the factors vanishes. Such a direction exists: their product is a nonzero polynomial, and a polynomial vanishing on all of $\mathbb R^n$ is zero. Use $\theta$ as the first coordinate direction. The leading $t$-coefficient of each factor is then its highest homogeneous part evaluated at $\theta$, independent of $w$. This change fixes the origin. No claim about weighted norms under this change is needed; we use it only for entire functions.

Lemma 3.1 (bounded removal). Let $H$ be a nonzero polynomial on $\mathbb C^d$. A holomorphic function on $\{H\ne0\}$ which is locally bounded on all of $\mathbb C^d$ extends holomorphically across $\{H=0\}$.

Proof. The constant case is immediate. Otherwise choose a coordinate direction in which $H$ has fixed positive degree and nonzero constant leading coefficient, as above. Write the coordinates $(t,w)$. Around a given point $(t_0,w_0)$, choose a small circle centered at $t_0$ on which $H(t,w_0)\ne0$. This is possible because that one-variable polynomial has finitely many zeros. The boundary remains free of zeros for all $w$ in a sufficiently small neighborhood of $w_0$.

On each such line, the function is bounded near the finitely many excluded points in the disk. The one-variable removable-singularity theorem extends it across those points. Inside the disk its extension equals $\frac{1}{2\pi i}\int_{|s-t_0|=R} \frac{h(s,w)}{s-t}\,ds. \tag{7}$ The integrand is holomorphic in the parameters on a neighborhood of the boundary, so this integral is jointly holomorphic in $(t,w)$. It agrees with $h$ wherever $H\ne0$. The local extensions agree on overlaps by continuity and the density of $\{H\ne0\}$, and thus give a global extension. $\square$

Lemma 3.2 (generic line division). Let $R(t,w)$ be a polynomial monic in $t$, and let $G$ be entire. If $G(t,w)/R(t,w)$ extends to an entire function of $t$ for every $w$ in a dense open subset of the parameter space, then $G/R$ is entire jointly in all variables.

Proof. Fix $(t_0,w_0)$, and choose a circle $|s-t_0|=R_0$ without zeros of $R(s,w_0)$. It stays free of zeros for nearby parameters. Define $J(t,w)$ inside the disk by $\frac{1}{2\pi i} \int_{|s-t_0|=R_0} \frac{G(s,w)}{R(s,w)(s-t)}\,ds. \tag{8}$ This is jointly holomorphic. For parameters in the dense open set, Cauchy's formula says that $RJ=G$ inside the disk. Continuity gives the same identity for all nearby parameters. Thus $J$ is a local extension of the quotient. These extensions agree off the polynomial zero set and therefore agree everywhere on overlaps. $\square$

Following the roots of an irreducible factor

The next argument uses global irreducibility, rather than irreducibility of an analytic germ.

Lemma 4.1 (connected simple-root cover). Let $r(t,w)$ be irreducible over $\mathbb C$, nonconstant, and monic in $t$. There is a nonzero polynomial $\Delta(w)$ such that every root of $r(t,w)$ is simple when $\Delta(w)\ne0$. The complex manifold $M_r=\{(t,w):r(t,w)=0,\ \Delta(w)\ne0\} \tag{9}$ is connected. If an entire function $G$ vanishes on a nonempty open subset of $M_r$, its restriction to $M_r$ vanishes identically.

Proof. When there are no parameters, irreducibility over $\mathbb C$ makes $r$ linear, and the assertion reduces to a single point. Otherwise Gauss's lemma makes $r$ irreducible in $\mathbb C(w)[t]$. In characteristic zero it is coprime to $\partial_t r$. A Bézout identity over that field, with denominators cleared, gives $\begin{aligned} A(t,w)r(t,w)&+B(t,w)\partial_t r(t,w)\\ &=\Delta(w),\qquad \Delta\ne0. \end{aligned} \tag{10}$ Hence no root is multiple off $\{\Delta=0\}$.

The base $U=\{\Delta\ne0\}$ is path connected. Indeed, for two points in $U$, the complex affine line joining them has a polynomial restriction of $\Delta$ which is nonzero at both endpoints. Only finitely many parameters on that line are excluded, and a path in the complex parameter plane avoids them. Locally over $U$, the simple roots are holomorphic functions, by the holomorphic implicit-function theorem. They continue along every path in $U$: on a compact path their values remain bounded by the monic root bound $\begin{gathered} |t|\le1+\max_{0\le a<q}|a_a(w)|,\\ r(t,w)=t^q+\sum_{a<q}a_a(w)t^a. \end{gathered} \tag{11}$ Local root charts can therefore be continued through a finite covering of the path. Continuation along loops permutes the roots above a base point.

Suppose these permutations have a proper nonempty orbit. The monic product of $t$ minus the roots in that orbit has single-valued holomorphic coefficients on $U$: different paths change the roots only by a permutation preserving the orbit. The coefficients are locally bounded near $\{\Delta=0\}$, since (11) bounds every elementary symmetric function of those roots. Lemma 3.1 extends each coefficient to an entire function of $w$. The coefficients also have polynomial growth, because the $a_a(w)$ are polynomials and (11) has a polynomial bound.

An entire function with $|h(w)|\le C(1+|w|)^N$ is a polynomial. The Cauchy estimates on polydisks of radius $R$ centered at zero give $|\partial^\alpha h(0)| \le C_\alpha(1+\sqrt d R)^N R^{-|\alpha|};$ letting $R\to\infty$ makes every derivative of order greater than $N$ zero. Its entire Taylor series is therefore finite. Apply this to the coefficients of the orbit product and of its complementary product. They produce a factorization of $r$ into two nonconstant polynomials, contradicting irreducibility. Thus the loop permutations act transitively.

Any two points of (9) can now be joined by a path: first lift a base path, and then use a lifted loop to reach the desired root in its fiber. This proves connectedness. The identity theorem in overlapping holomorphic root charts gives the final assertion. $\square$

Lemma 4.2 (propagating local division). Suppose every nonconstant irreducible factor of $R$ vanishes at zero. If $G$ is entire and $G/R$ is holomorphic near zero, then $G/R$ is entire.

Proof. Make all factors monic in $t$, and write $R=c\prod r_j^{m_j}$. Fix $j$. Since $r_j(0,0)=0$, a small circle centered at $t=0$ encloses this root and has no root on its boundary when $w=0$. For all sufficiently small $w$, there is still a root inside the circle, by the argument principle. Choose the circle and parameters small enough that their product neighborhood lies in the region of holomorphic division. There are parameters in this neighborhood with $\Delta_j(w)\ne0$. Thus a nonempty open root patch in $M_{r_j}$ lies in that region.

On this patch $G=RG_0$ for a holomorphic $G_0$. At the root of $r_j$, differentiation in $t$ gives $\partial_t^aG=0,\qquad 0\le a<m_j. \tag{12}$ Each function in (12) is entire, so Lemma 4.1 propagates its zero restriction throughout $M_{r_j}$.

There is a dense open set of parameters for which all roots of all the $r_j$ are simple and different factors have no root in common. To see the latter assertion, apply the Euclidean algorithm over $\mathbb C(w)$ to each distinct pair and clear denominators, just as in (10). The product of the finitely many resulting nonzero polynomials and the $\Delta_j$ is nonzero. Outside its zero set, (12) gives precisely the vanishing order $m_j$ needed at each root of $R(t,w)$. Thus $G/R$ is entire on every such $t$-line. Lemma 3.2 gives the joint extension. $\square$

Lemma 4.3 (division for a squarefree polynomial). If $R$ is squarefree and an entire function $G$ vanishes on $\{R=0\}$, then $G/R$ is entire.

Proof. In monic coordinates, outside the finite polynomial exceptional set constructed in the preceding proof, all roots of $R(t,w)$ are simple. Vanishing of $G$ removes each one-variable pole. Apply Lemma 3.2. A nonzero constant divisor needs no argument. $\square$

For an example of the local/global distinction, consider $r(t,w)=t^2-w^2(w+1). \tag{13}$ It is irreducible as a polynomial, but near $(0,0)$ it has the two analytic factors $t-w\sqrt{1+w}$ and $t+w\sqrt{1+w}$. The simple-root cover connects these branches by continuation around $w=-1$. Lemma 4.2 does not assume that the germ at the origin has just one branch.

Proving the density assertions

Proof of sufficiency in Theorem 1.1. Suppose first that every $r_j(0)=0$, and let $L$ annihilate all polynomial solutions. Its compact smooth restriction $\mu$ has this annihilation property. The local Fourier input makes $F/P(-\cdot)$ holomorphic near zero. The factors $r_j(-\cdot)$ are still irreducible and vanish at zero. Lemma 4.2 makes this quotient entire. The compact inverse and support identity (5) now give (6), so Hahn–Banach proves polynomial density.

Suppose next that $P$ is squarefree and $L$ annihilates (3). Then $\begin{gathered} F(-z)=\mu(e^{ix\cdot z})=0\\ \text{when }P(z)=0. \end{gathered} \tag{14}$ Lemma 4.3 makes $F(\zeta)/P(-\zeta)$ entire. Apply (5), (6), and separation again. These arguments work for uncountable $I$: a separating functional is controlled by finitely many seminorms, and the approximation assertion concerns neighborhoods defined by such finite lists. $\square$

For necessity we use smooth compact annihilators, which remain continuous even when the weights are too weak to control pointwise derivatives.

Lemma 5.1. If $\mu\in C_c^\infty(X)$, then $u\longmapsto u(\mu) \tag{15}$ is a continuous complex-linear functional on (1).

Proof. Choose $\chi\in C_c^\infty(X)$ equal to one near $\operatorname{supp}\mu$, and choose any one index $i$. Fourier duality and Hölder's inequality give $|u(\mu)|\le C\|\chi u\|_{p_i,k_i} \left\|\frac{\widehat\mu(-\xi)}{k_i(\xi)}\right\|_{L^{p_i'}}. \tag{16}$ The constant accommodates the fixed Fourier normalization. A moderate weight and its reciprocal grow at most polynomially, while $\widehat\mu$ is Schwartz. The last norm is finite, including $p_i=1$, when $p_i'=\infty$. This proves continuity. $\square$

Proof of necessity in Theorem 1.1. Choose an irreducible factor $r$ witnessing failure of the appropriate condition, and set $Q=P/r$. A nonconstant polynomial on $\mathbb C^n$ has a zero $z_0$: use monic coordinates and the fundamental theorem of algebra. Choose $\rho\in C_c^\infty(X)$ with $\int\rho=1$, and put $\begin{gathered} \varphi(x)=e^{-ix\cdot z_0}\rho(x),\\ \mu=Q(-D)\varphi. \end{gathered} \tag{17}$ Then $\widehat\varphi(-z_0)=1$, and $\frac{\widehat\mu(\zeta)}{P(-\zeta)} =\frac{\widehat\varphi(\zeta)}{r(-\zeta)} \tag{18}$ Equality (18) initially holds off the zero set of $P(-\cdot)$. Its right-hand side has no holomorphic extension at $-z_0$. If an entire extension existed, multiplication by $r(-\zeta)$ would give $\widehat\varphi$ everywhere by the identity theorem, contradicting its value there. The exponential-polynomial Fourier input therefore supplies a homogeneous exponential-polynomial solution $h$ with $\mu(h)\ne0$. Such an $h$ belongs to every space in (1), by smoothness. Lemma 5.1 says that $L(u)=u(\mu)$ is a continuous, nonzero functional on $\mathcal N_P(X)$.

For the polynomial criterion, assume $r(0)\ne0$. On the finite-dimensional space of polynomials of degree at most $N$, the operator $r(D)$ is invertible. Indeed, its constant part is $r(0)I$, and the remaining derivatives strictly lower degree, so their sum is nilpotent. A finite geometric series gives the inverse. If a polynomial $A$ satisfies $P(D)A=0$, the identity $r(D)Q(D)A=0$ therefore implies $Q(D)A=0$. Integration by parts in (17) yields $L(A)=\int\varphi Q(D)A=0$. The nonzero separating functional proves that polynomial solutions are not dense.

For the plane exponential criterion, assume that $r$ occurs at least twice in $P$. The polynomial $Q=P/r$ still contains every distinct irreducible factor of $P$. Thus $Q(z)=0$ at every zero of $P$, and $L(e^{ix\cdot z})=\int\varphi(x)Q(D)e^{ix\cdot z}\,dx=0.$ The same nonzero functional proves failure of density of their finite span. The construction used only a compact bump in $X$; it did not use convexity. $\square$

Repeated factors and displaced factors can occur together. For $P(z)=z^2(z-1)$ in one dimension, neither polynomial solutions alone nor plane exponentials alone are dense in the whole kernel. The three functions $1,x,e^{ix}$ are needed. The exponential-polynomial theorem includes all three without imposing either simplification criterion.

Exercises

Exercise 1 (introductory: degree-lowering inversion). Let $r(0)\ne0$. Write an explicit inverse for $r(D)$ on polynomials of degree at most $N$. Use it to show that every polynomial solution of $(D-2)^3u=0$ is zero.

Exercise 2 (intermediate: three approximation patterns). On a nonempty open interval, determine the full smooth kernels and the polynomial and plane exponential subspaces for $D^3$, $D(D-2)$, and $D^2(D-2)$. Relate them to Theorem 1.1.

Exercise 3 (advanced: local branches). Prove that (13) is irreducible in $\mathbb C[t,w]$, exhibit its two local analytic factors at zero, and explain why a loop around $w=-1$ connects their simple roots. Decide which approximation classes suffice for this symbol and for its square on an open convex subset of $\mathbb R^2$.

Exercise 4 (advanced: a weak topology still detects missing modes). Let $X$ be any nonempty open interval and equip solutions with the local $B_{2,(1+\xi^2)^{-4}}$ topology. Construct a smooth compact functional which annihilates every polynomial solution of $D(D-1)u=0$, but does not annihilate $e^{ix}$. Explain why evaluation of a derivative at a point is unnecessary.

Complete solutions

Solution 1. Put $T=r(0)^{-1}(r(D)-r(0)I)$. Each term of $T$ has at least one derivative, so $T^{N+1}=0$ on polynomials of degree at most $N$. Thus $r(D)^{-1}=r(0)^{-1}\sum_{j=0}^{N}(-T)^j.$ For $r(z)=(z-2)^3$, $r(0)=-8\ne0$. Every polynomial lies in some such finite-degree space, where the operator is invertible. Its polynomial kernel is zero. The full smooth kernel is nonzero; for example $e^{2ix}$ is a solution.

Solution 2. Since $D=-i\partial_x$, the kernel of $D^3$ is $\operatorname{span}\{1,x,x^2\}$. This follows by integrating $u'''=0$ three times. All are polynomial solutions, while the only plane exponential mode is $1$, because the only root of $z^3$ is zero. The origin criterion holds and the squarefree criterion fails.

For $D(D-2)$, set $v=(D-2)u$. Then $Dv=0$, so $v$ is constant. Solving the first-order equation gives $u=a+b e^{2ix}$. Its polynomial subspace is the constants: alternatively, $D-2$ is invertible on each finite-degree polynomial space, so $D(D-2)u=0$ implies $Du=0$. Both kernel generators are plane exponentials. The symbol is squarefree, and its factor $z-2$ does not vanish at zero.

For $D^2(D-2)$, $(D-2)u$ is affine. Since $D-2$ maps affine polynomials bijectively to affine polynomials, there is an affine particular solution and an added mode $e^{2ix}$. Hence the full kernel is $\operatorname{span}\{1,x,e^{2ix}\}$. Its polynomial subspace is $\operatorname{span}\{1,x\}$, and its plane exponential subspace is $\operatorname{span}\{1,e^{2ix}\}$. Both criteria fail. These finite-dimensional subspaces are closed in the Hausdorff solution topology, so the missing generators cannot be recovered by closure.

Solution 3. Regard (13) as a monic quadratic over $\mathbb C(w)$. Reducibility would mean that $w^2(w+1)$ is a square in that field. After division by the square $w^2$, this would make $w+1$ a square. A rational-function square has even order at each zero and pole, whereas $w+1$ has a zero of order one at $-1$. This is impossible. Gauss's lemma gives irreducibility in the polynomial ring.

The binomial power series for $\sqrt{1+w}$ converges near zero, and the two displayed analytic factors multiply to (13). Away from $w=0,-1$, the roots are $\pm w\sqrt{1+w}$. Along a small loop around $-1$, the argument of $1+w$ changes by $2\pi$; its square root changes sign, interchanging these roots. Joining that loop to a chosen base point gives the required continuation. Thus local splitting does not contradict the connected cover.

For $r$, the sole irreducible factor vanishes at zero and has multiplicity one. Both polynomial solutions and plane exponentials have dense spans. For $r^2$, polynomial solutions still have dense span, while plane exponentials do not, because the factor is repeated.

Solution 4. Choose $\rho\in C_c^\infty(X)$ with integral one and let $\varphi=e^{-ix}\rho$. Define $L(u)=u(-D\varphi).$ For smooth $u$, this equals $\int\varphi Du$. Polynomial solutions of $D(D-1)u=0$ are constants, so $L$ annihilates them. On the missing mode, $L(e^{ix})=\int e^{-ix}\rho(x)e^{ix}\,dx=1.$ The test function $-D\varphi$ is smooth and compactly supported. Its Fourier transform divided by $(1+\xi^2)^{-4}$ remains Schwartz, so (16) gives continuity in the specified topology. This compact integral functional detects the missing mode even though the norm does not control pointwise derivatives.

References

[Malgrange] B. Malgrange, Existence et approximation des solutions des équations aux dérivées partielles et des équations de convolution, Annales de l'Institut Fourier 6 (1956), 271–355. Original article and open PDF.