Euler chains and Fourier eigenspaces

Reconstructed and self-checked by GPT-6 Astra (OpenAI), Ultra reasoning effort, October 2026. Earlier edition: GPT-6.1 Sol (OpenAI), Ultra reasoning effort. Original exposition: CC0. Supplied foundations retain their stated licences.

Fourier transformation sends a generalized homogeneity parameter on the line to , without losing a chain. On Schwartz functions its symmetric normalization has four continuous projections. A scalar integrating factor then gives a global right inverse for each coordinate operator . For several such equations, we construct a solution exactly when the data satisfy the commuting-operator compatibility conditions.

For the Euler operator , a complex number , and a positive integer , write $Z(a,k)=\{u\in\mathcal D'(\mathbb R):(E-a)^ku=0\}. \tag{0.1}$ The complete proofs in U019, Theorems 2.1–2.2, give dimension , the exceptional negative-integer point jets, and the description on the two open half-lines, where each polynomial has degree below . Their coefficients need not be independently selectable at an exceptional parameter; the cited classification includes those constraints.

Our Schwartz and Fourier foundation, F1–F5, proves every seminorm estimate, compact-cutoff approximation, Gaussian constant, inverse identity and transposed coordinate identity used here. Its convention is $\begin{gathered} F\phi(\xi)=\int_{\mathbb R^n}e^{-ix\cdot\xi}\phi(x)\,dx,\qquad (Fu)(\psi)=u(F\psi),\\ F^2=(2\pi)^n\mathcal R,\qquad \mathcal R\phi(x)=\phi(-x). \end{gathered} \tag{0.2}$ Pairings are complex bilinear. We use , where . These are equivalent to F1's polynomial seminorms because , followed by the finite multinomial expansion. The scalar foundation, Sections 12.4–12.9, 13.1–13.5 and 13.7–13.10, supplies compactness, finite calculus, exponentials, logarithms and cutoffs. The integration foundation, Sections 15.0–15.1, supplies dominated convergence and absolute Fubini. Finite complex algebra is proved in the algebra foundation, Sections 10.1–10.3. These are supplied proofs, not external references in place of proofs.

Fourier duality includes every generalized Euler chain

Theorem 1.1 (Euler duality). Every member of has a unique tempered extension. With those extensions, $FZ(a,k)=Z(-a-1,k) \tag{1.1}$ is a linear bijection.

Proof: control the origin and infinity separately. Choose a compactly supported smooth equal to one near zero. On a fixed compact neighborhood of its support, continuity of gives integers and a constant with $|u(\psi)|\le C\max_{j\le r}\sup_K|\psi^{(j)}|, \qquad \operatorname{supp}\psi\subset K.$ Indeed a neighborhood of zero in this fixed-support test space imposes finitely many derivative bounds; scaling a test into that neighborhood gives the displayed inequality, with the largest of their orders. Consequently is defined for every , and $|u(\chi\phi)|\le C\max_{j\le r}\sum_{\ell=0}^j \binom j\ell\|\chi^{(j-\ell)}\|_\infty \|\phi^{(\ell)}\|_\infty \le C_\chi P_r(\phi).$

The distribution is the regular distribution of a smooth function , zero near zero, by U019's half-line classification. Choose an integer with . To check its growth explicitly, put for . For every nonnegative integer , is bounded: when its derivative changes sign at , and for it decreases. The finitely many logarithmic terms therefore give ; bounded intermediate intervals are covered by continuity. Hence $\left|\int b(x)\phi(x)\,dx\right| \le C P_{m+2}(\phi)\int_{\mathbb R}(1+x^2)^{-1}\,dx =C\pi P_{m+2}(\phi).$ The sum of the two pairings extends to . F1's compact-cutoff density proves uniqueness. This argument includes every exceptional point jet; its off-origin tail is simply zero.

Proof: transpose the operator. Multiplication by and differentiation are continuous on , by transposing their F1 bounds. Since vanishes on compact tests, density implies it vanishes on all Schwartz tests. F5 gives and . Thus the distributional product rule yields $\begin{aligned} F(Eu)&=i\partial_\xi(i\xi Fu)=-(E+1)Fu,\\ F((E-a)^ku)&=(-1)^k(E+a+1)^kFu. \end{aligned} \tag{1.2}$ Iteration is legitimate for each finite . This proves the inclusion in (1.1). Conversely, let , already tempered by the first part, and set . Applying (1.2) gives ; the inverse from F5 gives . This proves surjectivity, and the same inverse proves injectivity.

If , with , put . Each is in , so its transform exists. Equation (1.2) gives . When , also ; the normalized chain has exactly the same length, even at an exceptional degree.

Four finite projections give a unique decomposition

Theorem 2.1 (four eigenspaces). On , set . Then . Every has a unique sum , where , given by the continuous projections $u_k=\Pi_ku,\qquad \Pi_k=\frac14\sum_{j=0}^3i^{-kj}T^j. \tag{2.1}$

Proof. Equation (0.2) gives , and reflection twice is the identity. For an integer , write . If , . Otherwise , and , so the sum is zero. This proves the needed finite orthogonality without a spectral theorem.

Sum (2.1) over and use this identity to obtain . Replacing by its residue modulo four gives $T\Pi_k =\frac14\sum_{j=0}^3i^{-kj}T^{j+1} =i^k\Pi_k.$ If , then . Apply this to : . Every projection is a finite sum of continuous operators. The sum gives existence, and applying to any such decomposition gives uniqueness.

Write , with indices interpreted modulo four. Each eigenspace is closed: if an element has nonzero , some evaluation of that continuous function is nonzero, and continuity preserves that inequality in a neighborhood. This also explains the inherited Schwartz topology on .

A coordinate operator has a global Schwartz right inverse

Theorem 3.1 (coordinate surjectivity). For , let . Put and let contain the other coordinates. The formula $(R_\nu f)(s,z) =e^{-s^2/2}\int_0^s e^{t^2/2}f(t,z)\,dt \tag{3.1}$ defines a continuous linear map with . Moreover . For , is any complex constant.

Proof: smoothness and the zeroth -derivative. To put the finite integral on a fixed interval, substitute : $R_\nu f(s,z)=s\int_0^1 e^{-s^2(1-u^2)/2}f(su,z)\,du.$ Every parameter derivative of the integrand is continuous and bounded on compact parameter sets times . The fundamental theorem applied to difference quotients and dominated convergence justify differentiation of every order, including at . The original integral then gives .

For integers and a transverse multiindex , use $Q_{A,B,\beta}(f)= \sup_{t,z}\langle t\rangle^A\langle z\rangle^B |\partial_z^\beta f(t,z)|. \tag{3.2}$ Since each separate bracket is at most , . For , split the integral at . On its absolute value, after the outside Gaussian and , is at most $s e^{-3s^2/8}\langle z\rangle^{-B}Q_{0,B,\beta}(f).$ For , use and . The remaining exponential integral satisfies $e^{-s^2/2}\int_{s/2}^s e^{t^2/2}\,dt \le\int_{s/2}^s e^{-s(s-t)/2}\,dt \le\frac2s. \tag{3.3}$ The product is bounded: expand the integer bound and maximize each by differentiation. As , both pieces are controlled.

When , substitution gives the negative of the same integral over , with ; its absolute bound is unchanged. When , the length is at most two and between zero and . Enlarging the constant therefore proves the global estimate $\sup_{s,z}\langle s\rangle^A\langle z\rangle^B |\partial_z^\beta R_\nu f(s,z)| \le C_A Q_{A,B,\beta}(f). \tag{3.4}$

Proof: every mixed derivative and continuity. Define polynomials recursively by , with all nonexistent equal to zero, and $\begin{aligned} p_{r+1}&=p_r'-sp_r,\\ q_{r+1,\ell}&=q_{r,\ell}'+q_{r,\ell-1} +p_r\,\mathbf1_{\{\ell=0\}} \qquad(0\le\ell\le r). \end{aligned}$ Differentiating inductively proves $\partial_s^rR_\nu f =p_r(s)R_\nu f+ \sum_{\ell=0}^{r-1}q_{r,\ell}(s)\partial_s^\ell f. \tag{3.5}$ The recurrences give and ; for the latter, differentiation lowers degree, shifting the index adds the term of exactly the allowed degree, and has degree at most when at the next step. These assertions also hold for zero polynomials.

Choose constants bounding these finitely many polynomials by their indicated powers of . Apply (3.4) to with weight in the first term of (3.5). For the remaining terms use the definition of . This gives the explicit finite bound $\begin{split} \sup_{s,z}\langle s\rangle^A\langle z\rangle^B |\partial_s^r\partial_z^\beta R_\nu f| &\le C_{A,r}\left( Q_{A+r,B,\beta}(f) +\sum_{\ell<r} P_{A+B+r+|\beta|+\ell}(f)\right). \end{split}$ An empty sum is zero. Finally , because . Set and take the finite maximum over ; monotonicity of yields $P_N(R_\nu f)\le C_N P_{4N}(f)\qquad(N\ge0).$ This proves both Schwartz membership and continuity, with a finite input seminorm for each output seminorm.

Proof: the complete kernel. If , the integrating factor gives . For each , the scalar fundamental theorem makes that function constant in , so . Its trace is Schwartz, since .

Conversely, every derivative of is a polynomial times that Gaussian, by induction using . The same polynomial-exponential bound used above controls all its weights. Thus for any , each mixed derivative of has all separate weighted bounds, hence all joint bounds. This product lies in and is killed by . Subtracting from any solution now gives precisely this kernel.

The operator shifts the Fourier eigenvalue

The coordinate identities from F2 give $TL_\nu=iL_\nu T. \tag{4.1}$ Indeed and . Thus . The companion satisfies .

For , F3 proves , including the constant . Define and . Differentiation gives $A^m g_1=p_mg_1,\qquad T_1(p_mg_1)=(-i)^m p_mg_1.$ Induction shows that has leading coefficient , degree , and parity . For parity, and both have the next parity. The leading term cannot cancel a derivative of smaller degree, proving nonvanishing for every . No completeness assertion about Hermite expansions is needed.

For comparison, define the probabilists' Hermite polynomials by , . The chain rule and induction give . Thus the preceding proved identity is the normalization of NIST DLMF 18.17.22: its substitutions , turn the prefactor into , the Gaussian into , and the phase into . Parity changes to . The local recurrence proof supplies the whole identity used here.

Several coordinate equations require compatibility

Set . Absolute Fubini applies to products of Schwartz functions by F1's estimate. It factors the normalized Fourier transform into the product of the normalized one-dimensional transforms. In particular .

Theorem 4.2 (joint Schwartz system). For , the equations $L_\nu w=f_\nu,\qquad 1\le\nu\le n, \tag{J1}$ have a solution if and only if $L_\mu f_\nu=L_\nu f_\mu\quad\text{for every }\mu,\nu. \tag{J2}$ On the subspace of compatible tuples there is a continuous linear choice of solution, and all solutions differ by a scalar multiple of . If every datum lies in , there is a solution in . Within that eigenspace it is unique for , and has exactly the ambiguity for .

Proof: construct a solution by dimension induction. For distinct indices, expand . Coordinate multiplication commutes with the other coordinate derivative, and mixed derivatives commute for smooth functions, so . Equal indices are immediate. Applying these operators to (J1) proves necessity.

In dimension one choose . In larger dimension write , , and let $w_1=R_1f_1,\qquad a_\nu=f_\nu-L_\nu w_1\quad(\nu>1). \tag{J3}$ Using (J2) and , . Theorem 3.1 therefore gives $a_\nu(s,z)=e^{-s^2/2}h_\nu(z),\qquad h_\nu(z)=a_\nu(0,z)\in\mathcal S(\mathbb R^{n-1}). \tag{J4}$ For , $L_\mu a_\nu-L_\nu a_\mu =L_\mu f_\nu-L_\nu f_\mu -(L_\mu L_\nu-L_\nu L_\mu)w_1=0.$ The Gaussian depends only on . Substitute (J4) and set to obtain the same compatibility for the remaining operators acting on the . By induction choose satisfying for , and define $w(s,z)=w_1(s,z)+e^{-s^2/2}q(z). \tag{J5}$ This is Schwartz by the product estimate in Theorem 3.1. The added term is killed by ; for it contributes exactly . Thus all equations hold.

Proof: continuity and all ambiguities. Fix the just-described recursive choice of . The maps , , and the trace are continuous and linear by F1 and Theorem 3.1. The Gaussian product map is also continuous: for , its joint weighted derivative is bounded by . Induction and composition now give a continuous linear solution map on compatible tuples. This domain is closed in the product Schwartz topology: each compatibility difference is a continuous map into , whose zero set is closed by evaluation, and there are finitely many such differences.

For the joint kernel, the first equation gives ; the others give . Repeating the one-coordinate kernel computation in all remaining variables gives . Conversely for each , so every such difference occurs.

Proof: select the required eigenspace. Iterating (4.1) gives . Substitution in (2.1) yields $\begin{aligned} \Pi_{k+1}^{(n)}L_\nu &=\frac14\sum_{j=0}^3i^{-(k+1)j}i^jL_\nu T_n^j\\ &=L_\nu\Pi_k^{(n)}. \end{aligned} \tag{J6}$ If is the constructed solution, . This projects the entire compatible system at once. A difference in must also be , and its Fourier equation gives . This is the asserted exact uniqueness or Gaussian ambiguity.

Exercises

Exercise 1 (foundation). Find and on the line, retaining all constants. Check their Euler degrees.

Exercise 2 (intermediate). Given a normalized Euler chain of length three at , write its Fourier chain and check all relations and its length.

Exercise 3 (foundation). Compute and using reflection. Describe the even and odd Schwartz subspaces.

Exercise 4 (intermediate). For and , calculate , its Fourier eigenvalue and all its projections.

Exercise 5 (intermediate). Calculate and . Prove that each of the four Fourier eigenspaces on the line is nonzero.

Exercise 6 (foundation). Find every Schwartz solution of .

Exercise 7 (intermediate). Find every Schwartz solution on the plane of .

Exercise 8 (advanced). Show that is onto for every . Determine its kernel and which maps are bijective with continuous inverse.

Exercise 9 (advanced). In every dimension, give a continuous right inverse for and determine its entire kernel. Use , . When , exhibit a nonzero kernel element for each , and explain why the line behaves differently.

Exercise 10 (advanced). On the plane write . Determine all solutions and the indicated eigenspace restrictions for:

(a) , including the unique solution in ;

(b) , including every Fourier component and the exact meaning of the ambiguity;

Solutions

Solution 1. By definition , so . Inversion at zero gives , hence . Repeated use of the proved coordinate identities yields $F\delta_0^{(m)}=(i\xi)^m,\qquad F(x^m)=2\pi i^m\delta_0^{(m)}.$ For , testing on gives , so ; also . Therefore , whereas . These degrees obey .

Solution 2. Set , , . Theorem 1.1 permits all three transforms. Equation (1.2) gives $(E+a+1)w_0=0,\quad (E+a+1)w_1=Fv_0=w_0,\quad (E+a+1)w_2=-Fv_1=w_1.$ Since , invertibility implies . Thus the second iterated image of is nonzero and its third is zero: the length is exactly three.

Solution 3. In (2.1), the coefficients for add to zero when is odd and to when is even. Hence . Subtracting from gives . A function fixed by reflection is its even projection, so the even subspace is ; a function negated by reflection is its odd projection, so the odd subspace is . Directness follows by applying the individual .

Solution 4. The Gaussian identity in F3 is , so . Differentiation gives , hence . Since , . Theorem 2.1 gives and the other three projections zero. Nonvanishing follows, for example, by evaluation at .

Solution 5. The product rule gives . Starting with yields , and one more application yields . Their eigenvalues are respectively and . Together with and , these provide nonzero elements of all four eigenspaces: their polynomials have nonzero leading coefficients, and the Gaussian never vanishes.

Solution 6. Formula (3.1) gives . The full kernel gives exactly , . All these functions are Schwartz, and direct differentiation gives , .

Solution 7. The same integral gives the particular solution . Every solution, and no others, is $w(s,z)=e^{-s^2/2}\bigl(sz e^{-z^2}+h(z)\bigr), \qquad h\in\mathcal S(\mathbb R).$ The terms are Schwartz by the separate weighted derivative bounds, and differentiating in verifies the equation. Subtraction reduces any other solution to the whole kernel of Theorem 3.1.

Solution 8. For , let . Equation (J6) gives , proving surjectivity with a continuous right inverse. The unrestricted kernel is , entirely in . Its intersection with is therefore for , and zero for . The last three maps are bijections; their right inverse is necessarily their inverse and is continuous.

Solution 9. For every , define on $S_{\nu,k}f=\Pi_k^{(n)}R_\nu f,\qquad L_\nu S_{\nu,k}f=\Pi_{k+1}^{(n)}L_\nu R_\nu f=f. \tag{J7}$ This is continuous by Theorems 2.1 and 3.1. The unrestricted kernel is . Fubini and the Gaussian transform give, at frequency coordinates again denoted by , $T_nv(s,z)=g_1(s)\,(T_{n-1}h)(z). \tag{J8}$ Thus if and only if , since never vanishes. This proves the entire restricted kernel, including dimension one: gives , and .

For , choose a transverse coordinate and use the four already computed polynomials $p_0=1,\qquad p_1=2z_1,\qquad p_2=4z_1^2-2,\qquad p_3=8z_1^3-12z_1.$ Set $v_k(x)=p_m(z_1)g_n(x),\qquad m\in\{0,1,2,3\},\quad m\equiv-k\pmod4. \tag{J9}$ The factor has eigenvalue , and all other Gaussian factors have eigenvalue one; absolute Fubini multiplies these eigenvalues. Each is nonzero and , since its polynomial is independent of . All four restricted kernels are therefore nonzero. On the line there is no transverse factor, leaving only the Gaussian line in .

Solution 10. For any polynomial , cancellation of Gaussian derivatives gives and .

In (a), supplies both outputs. The joint kernel theorem gives all solutions . Each one-dimensional factor has eigenvalue , so lies in , while lies in . The projections are independent, so is exactly the unique solution. Both data have eigenvalue , agreeing with the shift.

In (b), supplies the two outputs. Again the full set is . The exact decomposition is $\begin{aligned} \tfrac12s^2&=\tfrac18(4s^2-2)+\tfrac14,\\ w&=\tfrac18(4s^2-2)g_2+(c+\tfrac14)g_2. \end{aligned} \tag{J10}$ The first component is nonzero in , and the second belongs to . Hence exactly gives a single-eigenspace solution, in ; every other solution has both components. There is no solution: its first output would lie in , whereas the nonzero datum lies in . The complete ambiguity refers to differences of unrestricted solutions, which are exactly , not to an solution for these data.

In (c), , violating (J2). Thus no common Schwartz solution exists. Individual coordinate surjectivity does not remove the proved compatibility condition for a tuple.

References

Michael E. Taylor, Fourier Analysis, Distributions, and Constant-Coefficient Linear PDE, free author notes, Section 3, formulas (3.11)–(3.24), PDF pages 26–28. The Gaussian differential-equation calculation is compared with F3's fully supplied mass and transform proof. The complex continuation and assertions on those pages are not inputs.
NIST, Digital Library of Mathematical Functions, 18.17.22, freely readable Hermite transform formula. The recurrence, parity and normalization needed here are proved above.
U019, Theorems 2.1–2.2, gives the full homogeneous and generalized Euler classification used in Theorem 1.1. F1–F5 gives the complete Schwartz and Fourier input. The linked scalar, algebra and integration components retain their own licences.