Approximation and global solvability from support geometry

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

A regular fundamental solution solves an equation with compact data. Data with unrestricted growth near the boundary need a further idea. We construct solutions on increasingly large regions and correct their differences by homogeneous solutions. The geometric condition that makes this correction possible is a uniform bound on the supports of compact test functions under the transposed operator.

The prerequisites are Local regularity, sharp embeddings, and compactness, Operator strength and local inverses, the Hahn–Banach separation theorem, and the Baire theorem for Fréchet spaces. Malgrange's approximation paper [Malgrange] and Hörmander's treatment of constant-coefficient equations [HII] are basic references. The compact-support Fourier statements used below have the full internal proofs specified in the next section.

Fix a nonzero complex polynomial \(P\), put \(D=-i\partial\), and use complex-linear distribution pairings. The formal transpose is

For a compactly supported distribution \(v\), \(\operatorname{ch}\operatorname{supp}v\) denotes the convex hull of its support.

Continuous functionals, test families and compact limits, Section 1, proves the locally convex separation used below. The complete-metric Baire theorem and the Fréchet results are Banach estimates, quotient spaces and compact parameter arguments, Sections 6 and 14.

The compact-support Fourier inputs

An exponential-polynomial solution is a function

Equivalently, \(P(D+z)A=0\). The polynomial factor records multiplicities; the class includes more than the plane exponentials for which \(P(z)=0\).

We use the following two entry results.

• For every compactly supported distribution \(v\),

In particular, a compactly supported homogeneous solution is zero.

• If \(\mu\) is compactly supported and annihilates every solution in (1), then there is a compactly supported \(v\) with

The solution is unique by (2). In Fourier–Laplace language, annihilation makes \(\widehat\mu(z)/P(-z)\) entire, and polynomial division preserves the growth estimate needed for compact support.

The ordinary support-hull equality in (2) is proved in Compact Fourier division and multiplicity-sensitive annihilators, Theorem CF4.1, and in Convex supports and convolution cancellation, Corollary 4.3. To apply the ordinary-derivative version of that corollary, substitute minus i times each polynomial variable. Empty support and nonzero scalar operators are included. Theorem CF5.1 of the compact Fourier lesson proves the exact annihilator criterion (3), with all polynomial multiplicities and the transpose sign. Its Theorems CF2.1 and CF3.2 prove the entire growth criterion and the division estimate, retaining the same convex carrier. The Fourier and Cauchy facts used in those proofs are linked there. Hörmander's distribution and Fourier treatment [HI] and Malgrange's paper [Malgrange] give the classical context.

Approximation in several regularity scales at once

Let

where the \(k_i\) are moderate and \(I\ne\varnothing\). The index set may be uncountable. The topology is generated by all cutoff seminorms \(\|\chi u\|_{p_i,k_i}\).

Smooth functions belong to every space in (4), and the map from \(C^\infty(X)\) into \(\mathcal F(X)\) is continuous. Indeed each fixed weighted cutoff seminorm is bounded by finitely many uniform derivatives on a compact set. For a compactly supported member of \(\mathcal F\), convolution with a smooth approximate identity converges in every seminorm of (4). This uses \(p_i<\infty\). It applies to an uncountable family because convergence is required separately in each seminorm, and a neighborhood uses only finitely many of them.

Lemma 2.1. Let \(L\) be a continuous linear functional on \(\mathcal F(X)\). Its restriction to smooth functions is a distribution \(\mu\) with compact support in \(X\). There is a compact \(K_L\subset X\) such that

whenever \(\chi\) equals one near \(K_L\). If \(P^tv=\mu\) with \(v\in\mathcal E'(X)\), then \(L(u)=0\) for every \(u\in\mathcal F(X)\) satisfying \(P(D)u=0\).

Proof. Continuity bounds \(|L(u)|\) by a finite sum of cutoff seminorms. The union of the supports of those cutoffs is a suitable \(K_L\), proving (5). The smooth restriction is continuous in the \(C^\infty\) topology and has support in \(K_L\), hence defines \(\mu\in\mathcal E'(X)\).

For the last assertion, choose \(\chi\in C_c^\infty(X)\) equal to one near \(K_L\cup\operatorname{supp}v\). Extend \(\chi u\) by zero, and mollify it to smooth functions \(u_\varepsilon\). They converge to \(\chi u\) in \(\mathcal F(X)\). For sufficiently small \(\varepsilon\), \(P(D)u_\varepsilon=0\) near \(\operatorname{supp}v\), since \(P(D)(\chi u)=0\) there and the mollifier has shrinking support. Consequently

Pass to the limit and use (5) to obtain \(L(u)=L(\chi u)=0\). This argument never pairs a distribution \(\mu\) directly with a nonsmooth \(u\). \(\square\)

Theorem 2.2. If \(X\) is convex, finite linear combinations of the solutions in (1) are dense in \(\{u\in\mathcal F(X):P(D)u=0\}\), with the topology inherited from \(\mathcal F(X)\).

Proof. Suppose a continuous \(L\) annihilates those exponential-polynomial solutions. Its smooth restriction \(\mu\in\mathcal E'(X)\) has the same annihilation property. Input (3) gives \(P^tv=\mu\) with \(v\) compactly supported in \(\mathbb R^n\). By (2),

The last inclusion holds because the convex hull of a compact subset of an open convex set is still a compact subset of that set. Lemma 2.1 now says that \(L\) annihilates every homogeneous solution in \(\mathcal F(X)\). Hahn–Banach separation proves the asserted density. The solution space is closed because \(\mathcal F\) embeds continuously into distributions and distributional differentiation is continuous. \(\square\)

Here density means simultaneous approximation in any finite list of the chosen cutoff norms. If that topology is not metrizable, the assertion does not automatically provide one approximating sequence for every solution.

A support condition for Runge approximation

For an open \(Y\), write \(\mathcal E'(\overline Y)\) for global compactly supported distributions whose support is contained in the closed set \(\overline Y\). Their supports may meet its boundary. This differs from \(\mathcal E'(Y)\).

Theorem 3.1. Let \(X_1\subset X_2\) be open. Assume that

Restrictions to \(X_1\) of homogeneous solutions in \(\mathcal F(X_2)\) are dense among homogeneous solutions in \(\mathcal F(X_1)\).

Proof. Let \(L\) annihilate all such restrictions. Its smooth restriction is \(\mu\in\mathcal E'(X_1)\). Every exponential-polynomial solution lies in \(\mathcal F(X_2)\), so (3) gives a global compact \(v\) satisfying \(P^tv=\mu\). We first show \(\operatorname{supp}v\subset\overline{X_2}\).

Choose a fundamental solution \(E^t\) of \(P^t\). Compact inversion gives \(v=E^t*\mu\). Let \(\check E^t(x)=E^t(-x)\). It is a fundamental solution of \(P(D)\). For \(h\in C_c^\infty(\mathbb R^n\setminus\overline{X_2})\), the convolution pairing identity yields

The function \(\check E^t*h\) is smooth and satisfies \(P(D)(\check E^t*h)=h=0\) on \(X_2\). Its restriction to \(X_2\) is therefore one of the solutions annihilated by \(L\), so the right side of (7) is zero. This proves the claimed closed-set support inclusion. Hypothesis (6) then improves it to \(v\in\mathcal E'(X_1)\). Lemma 2.1 and Hahn–Banach finish the proof. \(\square\)

The closure in (6) is essential to this proof: testing outside \(\overline{X_2}\) initially proves a closed-set inclusion, not compact support strictly inside \(X_2\).

Corollary 3.2. Suppose every global compact distribution \(v\) with \(\operatorname{supp}P^tv\subset X\) has support compactly contained in \(X\). Then exponential-polynomial solutions are dense among homogeneous solutions in \(\mathcal F(X)\).

Proof. Apply Theorem 3.1 with \(X_2=\mathbb R^n\), then apply Theorem 2.2 on \(\mathbb R^n\). Given finitely many seminorms and a tolerance, first approximate by a global solution and then approximate that solution by a finite exponential-polynomial sum, using half the tolerance at each step. \(\square\)

Convexity defined by supports

An open set \(X\) is \(P\)-convex for supports if for each compact \(K\subset X\) there is a compact \(K'\subset X\) such that

This is a property of the equation and the open set together. It asks for one support bound for all test functions with the indicated image support.

Every open convex set has this property: (2) allows \(K'=\operatorname{ch}K\). We next give a distance test which also handles nonconvex sets. Write

We also write \(d_X(x)=d_X(\{x\})\) for a point. The distance to the empty complement is \(+\infty\). We exclude the zero distribution when using distances of its support; it causes no condition in (8).

Theorem 4.1. An open set \(X\) is \(P\)-convex for supports if and only if

Proof. Suppose (9) holds. For nonempty compact \(K\subset X\), put \(\delta=d_X(K)>0\). If \(\operatorname{supp}P^t\phi\subset K\), (9) puts \(\operatorname{supp}\phi\) at distance at least \(\delta\) from the complement. By (2) it also lies in \(\operatorname{ch}K\). Thus one can use

a compact subset of \(X\). For \(X=\mathbb R^n\), simply use \(\operatorname{ch}K\); for empty \(K\), injectivity gives \(\phi=0\).

Conversely, assume (8), and first take nonzero \(v\in C_c^\infty(X)\) with finite \(\delta=d_X(\operatorname{supp}v)\). Suppose the right side of (9) were greater than \(\delta\). All translates \(v_a(x)=v(x-a)\), \(|a|<\delta\), belong to \(C_c^\infty(X)\). Their transposed images have supports in the single compact subset

By (8), all supports of \(v_a\) must lie in one compact subset of \(X\). But choose a nearest pair \(x\in\operatorname{supp}v\), \(y\notin X\), \(|y-x|=\delta\). Translating by \((1-1/j)(y-x)\) moves \(x\) arbitrarily close to \(y\), a contradiction. Such a nearest pair exists by compactness of the first support and closedness of the complement. The reverse inequality in (9) follows from locality, \(\operatorname{supp}P^tv\subset\operatorname{supp}v\). If the complement is empty, both distances are infinite.

For a general nonzero \(v\in\mathcal E'(X)\), mollify it. For small \(\varepsilon\), both \(v_\varepsilon\) and \(P^tv_\varepsilon\) have compact support in \(X\), and the preceding equality applies. Their support distances converge to the distances for \(v\) and \(P^tv\). Indeed convolution enlarges each original support by at most \(\varepsilon\), giving one bound. For the other, distributional convergence prevents the mollified functions from eventually vanishing on a neighborhood of a nearest support point. Input (2) ensures \(P^tv\ne0\). Passing to the limit proves (9). \(\square\)

Corollary 4.2. Interiors of arbitrary intersections of \(P\)-convex open sets are \(P\)-convex. Directed unions are \(P\)-convex, and so is

For an arbitrary indexed family, the same conclusion holds for the open lower limit

Every open set is contained in a smallest \(P\)-convex open set.

Proof. If \(X=\operatorname{int}\bigcap_iX_i\), its complement is the closure of the union of their complements. For \(v\in\mathcal E'(X)\), take the infimum over \(i\) of the equal distances in (9) for \(X_i\). This gives (9) for \(X\).

For a directed union \(X=\bigcup_iX_i\), each compact support lies in one \(X_i\): use a finite subcover and an upper member containing that finite subfamily. For all larger members the distance identities hold. Distances to their decreasing closed complements have supremum equal to the distance to their intersection, which is \(\mathbb R^n\setminus X\). To justify this when the supremum is finite, fix any larger radius. Each complement meets the compact closed neighborhood of the support with that radius. These intersections have the finite intersection property, so compactness supplies a point in every complement at that distance or less. Let the radius decrease to the supremum. If the supremum is infinite, the equality is immediate. Thus (9) passes to the union. Apply the intersection and directed-union conclusions to (11) and to the family indexed by finite exceptional sets \(J\).

Finally, take the interior of the intersection of all \(P\)-convex open sets containing the given open set. This family is nonempty because it contains \(\mathbb R^n\); its interior still contains the original open set. The intersection result proves the required minimality and convexity. \(\square\)

Why solvability forces this geometry

Theorem 5.1. If for every \(f\in C^\infty(X)\) there is a distributional solution of \(P(D)u=f\) in \(X\), then \(X\) is \(P\)-convex for supports.

Proof. Fix compact \(K\subset X\). Let \(G\) consist of the test functions \(\phi\in C_c^\infty(X)\) with \(\operatorname{supp}P^t\phi\subset K\), topologized by the increasing seminorms

It is metrizable and Hausdorff by compact injectivity. Put \(B(f,\phi)=\int f\phi\), for \(f\in C^\infty(X)\). For fixed \(\phi\), this is continuous in \(f\). For fixed \(f\), choose a solution \(u\). Then \(B(f,\phi)=u(P^t\phi)\), bounded by a constant times some \(q_N(\phi)\), because the argument of \(u\) is supported in the fixed \(K\).

Here is the needed joint-continuity argument. On the Fréchet space \(C^\infty(X)\), the closed sets

cover the space. By Baire, one \(A_l\) has interior. Taking differences of two points in that interior gives a neighborhood of zero on which \(|B(f,\phi)|\leq2lq_l(\phi)\). A smaller neighborhood is defined by a finite derivative seminorm \(p\) on a compact \(K'\subset X\). Scaling, and then passing to a limit if \(p(f)=0\), gives

Completeness of \(G\) was not needed.

If \(f\) is supported outside \(K'\), then \(p(f)=0\), so \(\int f\phi=0\) for all such \(f\). Therefore every \(\phi\in G\) has support in \(K'\). This is exactly (8). \(\square\)

The hypothesis asserts a solution for each datum. It does not assume a continuous or linear choice of solutions.

Correcting solutions on an exhaustion

We now prove the converse with precise weighted regularity. Suppose \(X\) is \(P\)-convex for supports, and let

Both spaces are Fréchet, and \(P(D):\mathcal G(X)\to\mathcal F(X)\) is continuous.

For integers \(\nu\geq1\), use the exhaustion

Then \(\overline{X_\nu}\subset X_{\nu+1}\), and the union is \(X\). Choose \(\phi_\nu\in C_c^\infty(X_\nu)\) equal to one near \(\overline{X_{\nu-1}}\), with \(X_0=\varnothing\). The seminorms

generate the topology of \(\mathcal G(X)\).

The exhaustion has the support property needed for Theorem 3.1. If \(v\in\mathcal E'(\overline{X_{\nu+2}})\) and \(\operatorname{supp}P^tv\subset X_\nu\), then \(v\in\mathcal E'(X)\). Identity (9) places its support at distance strictly greater than \(1/\nu\) from the complement. Identity (2) places it in the convex hull of the image support, and hence in a ball of radius strictly less than \(\nu\). Thus \(v\in\mathcal E'(X_\nu)\).

Lemma 6.1. If \(g\in\mathcal F(X)\) vanishes on \(X_\nu\), then for any \(\varepsilon>0\) there is \(w\in\mathcal G(X)\) with

Proof. The compact datum \(\phi_{\nu+2}g\), extended by zero, belongs globally to every \(B_{p_j,k_j}\). A regular fundamental solution gives one solution \(v=E*(\phi_{\nu+2}g)\) with all the local gains in (13). It is homogeneous on \(X_\nu\). By the support property just proved and Theorem 3.1, there is a homogeneous \(h\in\mathcal G(X_{\nu+2})\) such that

Take \(w=\phi_{\nu+2}(v-h)\), extending by zero. This lies in \(\mathcal G(X)\). On \(X_{\nu+1}\), the cutoff is identically one and \(P(D)v=g\), so the equation in (16) holds. On every \(\operatorname{supp}\phi_\mu\), \(\mu\leq\nu\), the same cutoff equals one, preserving the small seminorms. \(\square\)

Theorem 6.2. Every \(f\in\mathcal F(X)\) has a solution \(u\in\mathcal G(X)\) of \(P(D)u=f\).

Proof. First solve for compactly cut off data equal to \(f\) on \(X_1\), using a regular fundamental solution, and obtain \(u_1\in\mathcal G(X)\). Suppose \(u_\nu\) solves on \(X_\nu\). Its residual \(g_\nu=f-P(D)u_\nu\) belongs to \(\mathcal F(X)\) and vanishes on \(X_\nu\). Apply Lemma 6.1 with \(\varepsilon=2^{-\nu}\), and define \(u_{\nu+1}=u_\nu+w_\nu\). Then it solves on \(X_{\nu+1}\), and

For fixed \(\mu,j\), the tail of the sequence is Cauchy by summing the geometric series. Completeness gives a common \(u\in\mathcal G(X)\). Distributional continuity of \(P(D)\), applied on each exhaustion set, gives \(P(D)u=f\) throughout \(X\). \(\square\)

Every correction uses the same output weights \(k_jS_P\). The summation therefore retains the regularity gain; it does not accumulate a loss of derivatives at successive steps. The procedure is the linear-equation counterpart of the classical Mittag-Leffler construction.

Corollary 6.3. The following are equivalent:

1. \(X\) is \(P\)-convex for supports.
2. Every smooth datum has a smooth solution in \(X\).
3. Every smooth datum has a distributional solution in \(X\).

Proof. For (1) implies (2), apply Theorem 6.2 with \(p_j=2\) and \(k_j=\langle\xi\rangle^j\), \(j=0,1,\ldots\). Smooth data belong to all these local spaces. The solution belongs to all local polynomial Sobolev orders because \(S_P\) has a positive lower bound. The embedding criteria already proved in the local-regularity lesson make it smooth. Assertion (2) implies (3), and Theorem 5.1 proves (3) implies (1). \(\square\)

Corollary 6.4. Every distribution of finite order on \(X\) has a solution of finite order on \(X\) when \(X\) is \(P\)-convex for supports.

Here finite order means that one derivative order works in the test-function estimate on all compact subsets; the constants may depend on the subset. The order of the solution need not equal that of the datum.

Proof. If \(f\) has order \(N\), each \(\chi f\) has Fourier growth bounded by a constant times \(\langle\xi\rangle^N\). Hence \(f\in B_{1,\langle\xi\rangle^{-M}}^{\mathrm{loc}}\) for any integer \(M>N+n\). Theorem 6.2 supplies a solution in the same local space, since its output weight is at least a positive multiple of \(\langle\xi\rangle^{-M}\). Membership there gives an order at most \(M\) on every compact subset: the distribution pairing is bounded by the weighted Fourier \(L^1\) norm times \(\sup\langle\xi\rangle^M|\widehat\psi(\xi)|\), which is bounded by finitely many derivatives of \(\psi\) of order at most \(M\) on its fixed compact support. Thus one order works throughout \(X\). \(\square\)

An arbitrary distribution can have orders that grow along an exhaustion. Corollary 6.4 alone does not establish solvability for all distributions. The next two results describe its finite-order class exactly and show why hypoellipticity makes the additional singular-support condition automatic.

Which distributions fit a single weighted scale?

Write \(\mathcal D'_F(X)\) for distributions with one finite derivative order valid on every compact subset of \(X\), with constants allowed to depend on that subset.

Lemma 6.5. For every fixed \(p\in[1,\infty]\),

The order in the definition is uniform over the compact subsets, rather than a different order on each subset.

Proof. If \(f\) has order \(N\), each compact localization has the Fourier estimate

Indeed, pair \(f\) with a compact cutoff times the Fourier exponential and apply its order-\(N\) estimate. Every derivative of that exponential contributes at most one factor of the frequency; the cutoff derivatives have fixed bounds. For any integer \(M>N+n\), multiplying (FO2) by \(\langle\xi\rangle^{-M}\) gives an \(L^p\) function for every \(1\le p\le\infty\). This proves membership in the last union. Those weights are moderate: the inequality

applied also with \(-\eta\) bounds either quotient of the two shifted weights by a fixed power of \(\langle\eta\rangle\).

Conversely, let \(f\in B_{p,k}^{\rm loc}(X)\) for one moderate weight. Its defining shift bound, applied with one endpoint zero, gives constants \(c>0\), \(a\ge0\) such that

Fix a compact \(K\subset X\), choose \(\chi\) equal to one near \(K\), and let \(\psi\) be a smooth test supported in \(K\). Complex-linear Fourier duality and Hölder's inequality give

Both endpoint conventions are included. Choose an integer \(L>a+n\), independently of \(K\). Integration by parts in the Fourier integral gives

For example, bound each Fourier transform of a derivative by its \(L^1\) norm and use that \(\langle\xi\rangle^L\) is bounded by a constant times the sum of the coordinate monomials of orders at most \(L\). Every such \(L^1\) norm is bounded by a fixed compact-volume constant times the displayed maximum. By (FO3), the norm in (FO4) is at most that maximum times

The chosen \(L\) works for every \(p'\), including one and infinity. Thus

One derivative order works on all compact subsets, as required. This proves the reverse inclusion and (FO1). Empty domains and zero distributions satisfy all three descriptions trivially. \(\square\)

When hypoellipticity removes the order restriction

A constant-coefficient operator is hypoelliptic if, on every open set, a distribution with smooth image is itself smooth there.

Corollary 6.6. If \(P(D)\) is hypoelliptic and \(X\) is \(P\)-convex for supports, then every \(f\in\mathcal D'(X)\) has a solution \(u\in\mathcal D'(X)\). The datum need not have a uniform finite order.

Proof. Hypoellipticity passes to the transpose \(P^t=P(-D)\). If \(v\) is defined on an open set \(U\), reflect it to \(-U\) by

The chain rule, followed by distributional transposition, gives

Consequently a smooth \(P^t v\) makes \(Rv\), then \(v\), smooth. Locality gives the other inclusion between singular supports, so

For a compact \(v\) in \(X\), any compact bound \(K\) on the singular support of its image is therefore a bound on its own singular support. Thus \(X\) is \(P\)-convex for singular supports with \(K'=K\).

Theorem 6.1 of Singular supports and arbitrary distribution data proves that the two support-convexity conditions give a solution for every distributional datum. Apply that theorem. Its smooth solvability input is Corollary 6.3 above; no finite-order hypothesis on the present datum is used. \(\square\)

Exercises with solutions

Exercise 1 (entry). For \(P(\xi)=\xi^2\) in one variable, classify the solutions in (1). Why would using only plane exponentials lose a homogeneous solution?

Solution. A solution of \(D^2u=0\) is affine, so the exponential-polynomial solutions are \(a+bx\), with \(z=0\). One can also see this from \((D+z)^2A=0\): for \(z\ne0\), the highest nonzero coefficient of a polynomial \(A\) is multiplied by \(z^2\), so no nonzero \(A\) is possible; for \(z=0\), \(A\) is affine. Plane exponentials supply only constants, since zero is the only root. They cannot approximate \(x\) in the smooth topology on an interval. The polynomial factor captures the double root.

Exercise 2 (intermediate). On a nonempty open set, let \(P=c\ne0\) be constant. Identify its support-convexity condition, homogeneous approximation statement, and global inverse.

Solution. Multiplication by \(c\) preserves the support of every distribution, so every open set is \(P\)-convex and one may take \(K'=K\). The homogeneous solution space is zero, making both density theorems trivial. The global solution is \(u=f/c\). The constant weight \(S_P=|c|\) changes each weighted norm only by a fixed scalar, consistent with Theorem 6.2.

Exercise 3 (intermediate). Suppose an exhaustion correction sequence satisfies (17). For fixed \(\mu,j\), give a quantitative tail estimate once \(r\geq\max(\mu,j)\).

Solution. For \(s>r\), the triangle inequality gives

Pass to the limit to obtain the same bound for \(u-u_r\). This estimates one seminorm at a time. The finite initial segment has no effect on convergence.

Exercise 4 (advanced). Explain exactly which part of the proof fails if (4) includes \(p=\infty\). Does this prove that approximation or solvability is false at that endpoint?

Solution. Lemma 2.1 extends annihilation from smooth functions by mollifying a compact distribution and taking a limit in all the chosen norms. Approximate identities need not converge in \(B_{\infty,k}\), so that step is unavailable. The same extension is needed in the Runge argument and therefore in Lemma 6.1. This identifies a missing proof mechanism; it is not a counterexample to every possible endpoint theorem. The finite-exponent assumptions in the stated results cannot be dropped on the strength of this argument.

Exercise 5 (intermediate: an exact order threshold). Let \(n\ge1\), \(r\in\mathbb N_0\), and \(f=D_1^r\delta_0\) on \(\mathbb R^n\). For real \(M\), determine exactly when

Include \(p=1\) and \(p=\infty\).

Solution. The Fourier transform is \(\xi_1^r\). For \(1\le p<\infty\), its weighted \(p\)-th power is integrable near zero. Outside the unit ball, polar coordinates separate a positive finite angular factor from a radial integral comparable to

The angular factor is positive because \(|\theta_1|^{rp}\) is positive on an open part of the sphere; for \(r=0\), it is the sphere's area. The integral converges exactly when

For \(p=\infty\), the weighted function is bounded exactly when \(M\ge r\). Sufficiency follows from \(|\xi_1|\le|\xi|\le\langle\xi\rangle\); necessity follows along the first coordinate ray. The equality endpoint is therefore included only in this supremum case. The datum has a finite order independent of its localization, as Lemma 6.5 requires.

References

• [Malgrange] Bernard 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. See the preliminary scalar division result and Chapter I, Sections 2–3. Original article.
• [HI] Lars Hörmander, The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis, second edition, Springer, 2003 (reprint of the 1990 edition), Theorems 7.3.1–7.3.2 and Lemma 7.3.3, pp. 181–183; Definition 7.3.5 and Lemma 7.3.7, pp. 185–186. Publisher record.
• [HII] Lars Hörmander, The Analysis of Linear Partial Differential Operators II: Differential Operators with Constant Coefficients, Springer, 2005 (reprint of the 1983 edition), Sections 10.5–10.6, pp. 39–44: Theorems 10.5.1–10.5.2, Corollary 10.5.3, Definition 10.6.1, Theorems 10.6.3–10.6.4 and 10.6.6–10.6.7, Corollaries 10.6.5 and 10.6.8, Lemma 10.6.9, and Corollary 10.6.10. Publisher record.