Regular polynomial field functional algebra

Define $d_X:X\to X\times X$ by $d_X(x)=(x,x)$. For $\mathrm{\Psi }\in \mathcal{D}(X\times X)$, the diagonal restriction $x\mapsto \mathrm{\Psi }(d_X(x))$ has first derivatives ${\mathrm{\partial }}_{x_j}\mathrm{\Psi }(x,x)+{\mathrm{\partial }}_{y_j}\mathrm{\Psi }(x,x)$ by [F2]. Iterating this formula on the smooth partial derivatives of $\mathrm{\Psi }$ shows the restriction has continuous derivatives of every order, so it is smooth. Its support is contained in the first-coordinate projection of $\mathrm{supp}\mathrm{\Psi }$, which is compact as a continuous image of a compact set. The componentwise convention in the Statement and [F3] make the displayed integral well defined.

The assignment is complex-linear. Fix a nonempty compact $K\subseteq X\times X$ and a compact rectangle $R\subseteq X$ containing its first-coordinate projection. For every $\mathrm{\Psi }$ supported in $K$, $\mathrm{\mid }⟨{\delta }_{{\mathrm{\Delta }}_X},\mathrm{\Psi }⟩\mathrm{\mid }\le \mathrm{vol}(R){\sup }_{(x,y)\in K}\mathrm{\mid }\mathrm{\Psi }(x,y)\mathrm{\mid }\mathrm{.}$ Thus the seminorm $q(\mathrm{\Psi })=\mathrm{\mid }⟨{\delta }_{{\mathrm{\Delta }}_X},\mathrm{\Psi }⟩\mathrm{\mid }$ is continuous on every fixed-support test space. By the LF topology definition in [F1], such seminorms are continuous on $\mathcal{D}(X\times X)$. Therefore ${\delta }_{{\mathrm{\Delta }}_X}\in {\mathcal{D}}^{\mathrm{\prime }}(X\times X)$.

If an open set $U\subseteq X\times X$ is disjoint from ${\mathrm{\Delta }}_X$, then every test supported in $U$ vanishes at every $(x,x)$. Its pairing with ${\delta }_{{\mathrm{\Delta }}_X}$ is zero, so [F5] gives $\mathrm{supp}{\delta }_{{\mathrm{\Delta }}_X}\subseteq {\mathrm{\Delta }}_X$.

Fix $(x_0,x_0)\in {\mathrm{\Delta }}_X$ and any open neighborhood $U$ of this point. There is a ball $B$ about $x_0$ with $\bar{B}\times \bar{B}\subseteq U$. By [F4] choose $h\in \mathcal{D}(X)$ with $h\ge 0$, $h(x_0)=1$, and support in $B$. By continuity choose a closed nondegenerate cube $Q$ inside $B$ on which $h\ge 1\mathrm{/}2$, and choose a sufficiently large cube $R$ containing $\mathrm{supp}h$ and $Q$ with the faces of $Q$ among its grid cuts. The lower Darboux sum of the nonnegative function $h^2$ on this grid is at least $\mathrm{vol}(Q)\mathrm{/}4>0$, so its real Riemann integral is positive. The product test $\mathrm{\Psi }(x,y)=h(x)h(y)$ is supported in $U$, and $⟨{\delta }_{{\mathrm{\Delta }}_X},\mathrm{\Psi }⟩={\int }_{X}h(x{)}^2{d}^{4}x>0$. Thus ${\delta }_{{\mathrm{\Delta }}_X}$ does not vanish on any neighborhood of $(x_0,x_0)$, so $(x_0,x_0)\in \mathrm{supp}{\delta }_{{\mathrm{\Delta }}_X}$ by [F5]. Since the point was arbitrary, ${\mathrm{\Delta }}_X\subseteq \mathrm{supp}{\delta }_{{\mathrm{\Delta }}_X}$. Together with step 3.1 this gives equality.

For every $\mathrm{\Psi }$, the diagonal restrictions of $\mathrm{\Psi }(x,y)$ and $\mathrm{\Psi }(y,x)$ coincide, so the distribution is symmetric. The diagonal has empty interior: every product neighborhood of a point $(x,x)$ contains $(x,y)$ with $y\mathrm{\ne }x$. If a smooth function supported on the diagonal were nonzero at a point, continuity would make it nonzero on an open neighborhood, contradicting empty interior. Therefore every such smooth function is zero.

If $a=b$, every integral is zero, $U_0=I$, and $U_n=0$ for $n\ge 1$; all evolution and unitarity claims follow, while no $c<d$ lies in $I$, so the separated-support clause has no instance. Assume henceforth $a<b$. Suppose $U_n(r,s)\psi$ is continuous in $r$. For fixed $r_0$, $\mathrm{\parallel }V(r)U_n(r,s)\psi -V(r_0)U_n(r_0,s)\psi \mathrm{\parallel }\le M\mathrm{\parallel }{U}_n(r,s)\psi -U_n(r_0,s)\psi \mathrm{\parallel }+\mathrm{\parallel }(V(r)-V(r_0))U_n(r_0,s)\psi \mathrm{\parallel },$ which tends to zero by strong continuity of $V$. Thus the recursive integrand is continuous. Apply [F7] to $I=[a,b]$ and the Hilbert space with their norm metrics; the estimate gives uniform continuity of the integrand. On any interval with endpoints $s,t$, the midpoint tagged sums of its equal-mesh partitions are Cauchy by uniform continuity, so completeness of $\mathcal{H}$ gives a norm limit. The same modulus-of-continuity estimate shows that every tagged sum of sufficiently small mesh approaches this limit, proving existence and tag/partition independence of the Riemann integral defined in the Statement; reverse its sign when $t<s$. The tagged-sum triangle inequality and scalar Riemann-sum limit [F5] give $\mathrm{\parallel }\int f\mathrm{\parallel }\le \int \mathrm{\parallel }f\mathrm{\parallel }$. The integral as a function of its upper endpoint is continuous, since its increment has norm at most the interval length times the supremum of the continuous integrand; this supremum is finite by [F3]. This proves strong continuity of $U_{n+1}(\cdot ,s)\psi$ and closes the induction.

The identity operator has norm at most one, including when $\mathcal{H}=\{0\}$. The integral estimate then gives inductively $\mathrm{\parallel }{U}_{n+1}(t,s)\mathrm{\parallel }\le {\int }_{\min (s,t)}^{\max (s,t)}M\frac{M^n\mathrm{\mid }r-s{\mathrm{\mid }}^n}{n!}dr=\frac{M^{n+1}\mathrm{\mid }t-s{\mathrm{\mid }}^{n+1}}{(n+1)!}\mathrm{.}$ This proves the factorial bound for both orientations of the integral.

The bound in step 2.1 makes $\sum z^n{U}_n(t,s)$ absolutely and uniformly convergent in operator norm on $I^2$, with total norm at most ${\sum }_{n\ge 0}(\mathrm{\mid }z\mathrm{\mid }M(b-a){)}^n\mathrm{/}n!=e^{\mathrm{\mid }z\mathrm{\mid }M(b-a)}$ by [F4]. The recursion and this uniform bound permit termwise integration on each vector: the norm of the integral of a uniform tail is at most $(b-a)$ times its supremum norm, which tends to zero. This gives the displayed integral equation.

If $X(t,s)$ and $Y(t,s)$ are two strongly continuous solutions, then for fixed $\psi$, $D(t)=\mathrm{\parallel }(X(t,s)-Y(t,s))\psi \mathrm{\parallel }$ is bounded on $I$ by [F3] and is Riemann integrable by [F5]. It satisfies $D(t)\le \mathrm{\mid }z\mathrm{\mid }M{\int }_{\min (s,t)}^{\max (s,t)}D(r)dr$. Repeated substitution gives an $n$-fold integral over the ordered simplex between $s$ and $t$; induction using [F5] and the polynomial derivative rule [F6] evaluates its volume as $\mathrm{\mid }t-s{\mathrm{\mid }}^n\mathrm{/}n!$, so $D(t)\le C_{\psi }\frac{(\mathrm{\mid }z\mathrm{\mid }M\mathrm{\mid }t-s\mathrm{\mid }{)}^n}{n!}$ with $C_{\psi }={\sup }_{r\in I}D(r)<\mathrm{\infty }$. The right side tends to zero, so the solution is unique. To prove composition, split the equation for $U_z(t,s)$ at $r$: for $t\ge r$, it reads $U_z(t,s)=U_z(r,s)-iz{\int }_r^{t}V(q)U_z(q,s)dq$. Multiplying the equation for $U_z(t,r)$ on the right by $U_z(r,s)$ shows $U_z(t,r)U_z(r,s)$ solves this same initial-value equation and has the same value at $t=r$. The uniqueness estimate just proved, with initial time $r$, gives equality; the reversed orientation proves it for $t<r$.

For a continuous vector-valued integrand, the average over a signed interval of length $h$ converges in norm to its value at the base point: the error is bounded by the supremum of its variation on that interval. Taking difference quotients in the integral equation therefore gives ${\mathrm{\partial }}_t{U}_z(t,s)\psi =-izV(t)U_z(t,s)\psi$, with relative one-sided derivatives at the endpoints. If $u(t)=U_z(t,s)\psi$, then $\frac{\mathrm{\parallel }u(t+h){\mathrm{\parallel }}^2-\mathrm{\parallel }u(t){\mathrm{\parallel }}^2}{h}=2\mathrm{Re}⟨u(t),\frac{u(t+h)-u(t)}{h}⟩+\frac{\mathrm{\parallel }u(t+h)-u(t){\mathrm{\parallel }}^2}{h}\mathrm{.}$ The second term tends to zero and the first tends to $2\mathrm{Re}⟨u(t),-izV(t)u(t)⟩=0$ for real $z$, because self-adjointness makes $⟨u,V(t)u⟩$ real. By the continuity estimate in step 1.1, the scalar function $t\mapsto \mathrm{\parallel }u(t){\mathrm{\parallel }}^2$ has continuous derivative zero. The real fundamental theorem [F5] on each subinterval therefore makes $\mathrm{\parallel }u(t){\mathrm{\parallel }}^2$ constant, so $U_z(t,s)$ is an isometry.

By the composition law in step 4.1, $U_z(t,s)U_z(s,t)=U_z(t,t)=I_{\mathcal{H}}$ and $U_z(s,t)U_z(t,s)=U_z(s,s)=I_{\mathcal{H}}$. Hence $U_z(t,s)$ is surjective. Step 4.2 shows that it is an isometry when $z$ is real. By the unitary convention stated above, $U_z(t,s)$ is unitary.

Let $Y(t)=U_z^{V_1}(t,a)U_z^{V_2}(t,a)$. On $[a,c]$, $U_z^{V_1}(t,a)=I$, so the integral equation for $U_z^{V_2}$ is the equation for $Y$ with generator $V_1+V_2$. On $[c,d]$, both generators vanish on the open gap and, by strong continuity, at its endpoints; thus $Y(t)=U_z^{V_2}(c,a)$ is constant and the same equation continues across the gap. On $[d,b]$, $U_z^{V_2}(t,a)=U_z^{V_2}(c,a)$, while the integral equation for $U_z^{V_1}$ reduces to $U_z^{V_1}(t,a)=I-iz{\int }_d^t{V}_1(q)U_z^{V_1}(q,a)dq\mathrm{.}$ Right multiplication by the fixed bounded operator $U_z^{V_2}(c,a)$ commutes with the vector integral by its tagged-sum definition, so $Y$ satisfies the equation for $V_1+V_2$ on this interval too. The three interval equations agree at their endpoints and combine to the full integral equation on $I$; uniqueness from step 4.1 gives the stated factorization at $b$. Both sides have operator-norm-convergent power series for every $z$. Their equality for real $z$ forces equality of coefficients: a first differing coefficient would remain nonzero after division by its least power of $z$ and passage to $z\to 0$.

Fix $\mathrm{\Phi }\in \mathcal{D}(X^n)$ and regard $X^n$ as ${\mathbb{R}}^D$, $D=dn$. The coordinate projections of $\mathrm{supp}\mathrm{\Phi }$ lie in a compact $K\subset X$. By [F3] choose $\chi \in \mathcal{D}(X)$ equal to one near $K$. Choose $L>0$ so large that every coordinate of $\mathrm{supp}\chi$ lies strictly inside $(-L\mathrm{/}2,L\mathrm{/}2)$. Extend $\mathrm{\Phi }$ by zero and then $L$-periodically in each of the $D$ coordinates; because its support is interior to the period cube, the resulting function is smooth on the compact $D$-torus.

For $k\in {\mathbb{Z}}^D$, put $e_k(x)=\exp (2\pi ik\cdot x\mathrm{/}L)$, $Q=[-L\mathrm{/}2,L\mathrm{/}2{]}^D$, and $\widehat{\mathrm{\Phi }}(k)=L^{-D}{\int }_Q\mathrm{\Phi }(x)\overline{e_k(x)}dx$. These are Riemann integrals by [F7]. Periodic integration by parts in one coordinate has no boundary term: the endpoint product agrees by periodicity, and [F8] converts the integral of a derivative to that endpoint difference. Writing $\mathrm{\Delta }={\sum }_{j=1}^D{\mathrm{\partial }}_j^2$, applying this identity to $(1-\mathrm{\Delta }{)}^q\mathrm{\Phi }$, and using the character derivative from [F6], gives $\widehat{(1-\mathrm{\Delta }{)}^q\mathrm{\Phi }}(k)=(1+(2\pi \mathrm{/}L{)}^2\mathrm{\mid }k{\mathrm{\mid }}^2{)}^q\widehat{\mathrm{\Phi }}(k)$. The left side is bounded by the volume of $Q$ times the supremum of the continuous function $(1-\mathrm{\Delta }{)}^q\mathrm{\Phi }$, so $\mathrm{\mid }\widehat{\mathrm{\Phi }}(k)\mathrm{\mid }\le C_{\mathrm{\Phi },q}(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{-q}\mathrm{.}$ For integer $m\ge 1$, the number of lattice points in $m\le \mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}<m+1$ is at most $(2m+3{)}^D-(2m-1{)}^D\le C_D(m+1{)}^{D-1}$, by expanding the two powers; the shell $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}<1$ is finite. Thus, for every multi-index $\alpha$, choosing $q$ with $2q>D+\mathrm{\mid }\alpha \mathrm{\mid }$ makes ${\sum }_k\mathrm{\mid }k{\mathrm{\mid }}^{\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }\widehat{\mathrm{\Phi }}(k)\mathrm{\mid }$ finite. The Fourier series and each derivative series therefore converge absolutely and uniformly on the torus. Integrating each uniformly convergent derivative series along a coordinate segment and using [F8] shows it is the corresponding derivative of the uniformly convergent Fourier sum; iterating over coordinates gives the same assertion for every multi-index.

The uniform sum has the same Fourier coefficients as $\mathrm{\Phi }$, since uniform convergence allows integration term by term and distinct integer characters have zero cross integral by [F6]–[F8]. Their difference is a continuous $D$-periodic function with every multi-index Fourier coefficient zero. For $D=1$, all its Fejer means are finite sums of those zero coefficients, so [F5] makes the function zero. For $D>1$, fix the first frequency $k_1$ and integrate in the first coordinate to obtain a continuous $(D-1)$-periodic coefficient function; continuity follows from [F4] and the compact integral bound [F7]. Its remaining Fourier coefficients are the corresponding coefficients of the original difference by [F7], so induction makes each such coefficient function zero. Applying the one-dimensional case to each fixed choice of the remaining coordinates shows the original difference is zero. Hence the Fourier series sums to $\mathrm{\Phi }$, with all derivatives uniformly.

Let $S_M\mathrm{\Phi }$ be the finite Fourier sum over $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}\le M$. On ${\mathbb{R}}^D$, put $P_M(x_1,\dots ,x_n)=({\prod }_{j=1}^n\chi (x_j))S_M\mathrm{\Phi }(x_1,\dots ,x_n)\mathrm{.}$ Each Fourier character factors over the $n$ blocks $x_j\in {\mathbb{R}}^d$; after multiplication by the product cutoff, each term is an elementary tensor of tests in $\mathcal{D}(X)$. Also $({\prod }_j\chi (x_j))\mathrm{\Phi }=\mathrm{\Phi }$, since every coordinate projection of its support lies in $K$. The product rule [F8] and the derivative convergence from step 2.1 show $P_M\to \mathrm{\Phi }$ uniformly with every derivative, with all supports in the fixed compact $(\mathrm{supp}\chi {)}^n$. By [F1] this is convergence in the LF topology. Applying the finite symmetrization operator preserves convergence and turns the approximants into symmetric finite tensor sums whenever $\mathrm{\Phi }$ is symmetric.

Suppose ${\sum }_{n=0}^N\int f_n(x_1,\dots ,x_n){\prod }_j\varphi (x_j)dx=0$ for every real smooth $\varphi$, with symmetric test kernels $f_n$ as in [F9]. Replacing $\varphi$ by $t\varphi$ makes a scalar polynomial in $t$ vanish for every real $t$, so each homogeneous term $\int f_n{\varphi }^{\otimes n}$ is zero for every $\varphi$. For fixed real test functions $h_1,\dots ,h_n$, substitute $\varphi ={\sum }_j{t}_j{h}_j$. The coefficient of $t_1\cdots t_n$ is $n!\int f_n(x_1,\dots ,x_n){\prod }_j{h}_j(x_j)dx$, by symmetry, and must vanish. Complex product tests are finite linear combinations of real product tests by separating real and imaginary parts. Step 3.1 makes their span dense, so choose approximants ${\psi }_M\to \overline{f_n}$ with support in a fixed compact rectangle $C$. The componentwise convention [F7, F9] and the Riemann-sum triangle inequality give $\mid \int f_n({\psi }_M-\overline{f_n})\mid \le \mathrm{vol}(C)\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{\psi }_M-\overline{f_n}{\mathrm{\parallel }}_{\mathrm{\infty }}⟶0.$ Thus $\int \mathrm{\mid }{f}_n{\mathrm{\mid }}^2=0$. If $f_n$ were nonzero at some point, its support constraint and continuity give a closed nondegenerate rectangle $Q\subset \mathrm{int}C$ and $c>0$ with $\mathrm{\mid }{f}_n\mathrm{\mid }\ge c$ on $Q$. Insert the faces of $Q$ into a Darboux grid of $C$. Since $\mathrm{\mid }{f}_n{\mathrm{\mid }}^2\ge 0$ everywhere and $\mathrm{\mid }{f}_n{\mathrm{\mid }}^2\ge c^2$ on $Q$, the lower Darboux sum is at least $c^2\mathrm{vol}(Q)>0$, so the real Riemann integral is positive, a contradiction. Hence $f_n=0$. The constant term is zero by taking $\varphi =0$. This proves uniqueness and the well-definedness claim used by the definition.

For each $k$, [F2] constructs $W^{\otimes k}$ on the independent variable pairs. The kernels $F^{(k)}(\varphi )$ and $G^{(k)}(\varphi )$ in [F1] are smooth and compactly supported, so the displayed pairing is defined. In each homogeneous term of $F$ and $G$, pairing the $k$ contracted variables leaves a smooth compact kernel in the remaining variables: the support lies in the projection of the original compact coefficient supports, and [F3] passes every remaining-variable derivative through the pairing. The raw remaining-variable kernel need not be symmetric under permutations exchanging a variable from the $F$ block with one from the $G$ block. For each output degree $r$, replace it by its full symmetrization ${\mathrm{Sym}}_{r}h=\frac{1}{r!}{\sum }_{\pi \in S_r}h(x_{\pi (1)},\dots ,x_{\pi (r)})$. This preserves smoothness and compact support, and its field evaluation is unchanged because both the product ${\prod }_{j=1}^r\varphi (x_j)$ and the product integration measure are invariant under permutations. The resulting symmetric coefficient therefore represents each summand in ${\mathcal{P}}_{\mathrm{r}\mathrm{e}\mathrm{g}}(X)$.

If $F$ and $G$ have degrees $m$ and $n$, then $F^{(k)}=0$ for $k>m$ and $G^{(k)}=0$ for $k>n$. Hence only $0\le k\le \min (m,n)$ contribute; the sum is finite, independent of a coefficient presentation by [F1], and belongs to ${\mathcal{P}}_{\mathrm{r}\mathrm{e}\mathrm{g}}(X)[\eta ]$.

For $f\in \mathcal{D}(X)$, define the auxiliary formal generator by $E_f(s,\varphi ):={\sum }_{m\ge 0}{s}^m{\mathrm{\ell }}_f(\varphi {)}^m\mathrm{/}m!$ in the formal power-series ring; it is not asserted to be an element of ${\mathcal{P}}_{\mathrm{r}\mathrm{e}\mathrm{g}}(X)$. For a zero-constant formal series $Z$, $\exp Z$ means ${\sum }_{r\ge 0}{Z}^r\mathrm{/}r!$, with all equalities interpreted coefficientwise. The $k$-th derivative of $E_f$ in $\varphi$ is $s^k{f}^{\otimes k}{E}_f$. Substitution into the finite coefficient formula gives, coefficientwise in $s,t,\eta$, $E_f(s){\star }_W{E}_g(t)=\exp \text{\hspace{0.17em}⁣}(\eta stW(f,g))E_{sf+tg}\mathrm{.}$

Applying step 2.2 twice, both bracketings of three generators equal $\exp \text{\hspace{0.17em}⁣}(\eta (stW(f,g)+srW(f,h)+trW(g,h)))E_{sf+tg+rh},$ by bilinearity of $W$. Comparing coefficients of $s,t,r$ proves associativity for powers of linear functionals. For distinct tests in one input, replace $f$ by ${\sum }_{j=1}^m{s}_j{f}_j$ in the same coefficientwise identity; its coefficient of $s_1\cdots s_m$ is the symmetrized product ${\mathrm{\ell }}_{f_1}\cdots {\mathrm{\ell }}_{f_m}$. Make this substitution independently in each of the three inputs. Thus the associator vanishes for every symmetrized elementary tensor-product coefficient kernel and for their finite sums.

By [F1], symmetrized finite product-test kernels are dense in every coefficient test space. Each fixed coefficient of the associator is a finite sum of compositions of continuous tensor-distribution pairings from [F2] and smooth parameter pairings from [F3], hence is continuous in the coefficient kernels. Passing to the limits of the product-test approximants from the preceding coefficient comparison proves associativity for arbitrary $F,G,H\in {\mathcal{P}}_{\mathrm{r}\mathrm{e}\mathrm{g}}(X)$.

For linear functionals only $k=0,1$ occur. The $k=0$ term cancels in the commutator, while the $k=1$ terms are $\eta W(f,g)$ and $\eta W(g,f)$; this gives the displayed commutator. If $W$ is Hermitian, conjugating each finite contraction reverses its two variable groups and gives $(F{\star }_{W}G{)}^{\ast }=G^{\ast }{\star }_W{F}^{\ast }$.

For a homogeneous coefficient of degree $m$, one application of ${\mathrm{\Gamma }}_K$ contracts two derivative variables against $K$, leaving a smooth compact kernel by [F1]–[F3]. It lowers the polynomial degree by two; therefore ${\mathrm{\Gamma }}_K^{j}F=0$ when $2j>\mathrm{deg}F$, so ${\alpha }_{K}F$ is a finite sum in ${\mathcal{P}}_{\mathrm{r}\mathrm{e}\mathrm{g}}(X)[\eta ]$. The same finite argument defines ${\alpha }_{-K}$.

For this computation set $E_f(s,\varphi ):={\sum }_{m\ge 0}{s}^m{\mathrm{\ell }}_f(\varphi {)}^m\mathrm{/}m!$ as a formal generating series. Each second derivative contracts to $K(f,f)E_f$. Coefficientwise this gives ${\alpha }_K{E}_f(s)=\exp \text{\hspace{0.17em}⁣}(\frac{\eta s^2}{2}K(f,f))E_f(s),$ where the exponential is its formal power series; each coefficient used below involves only finitely many derivatives.

On two auxiliary generators, the left side of the asserted identity has the scalar factor $\exp \text{\hspace{0.17em}⁣}(\eta stW(f,g)+\frac{\eta }{2}K(sf+tg,sf+tg))\mathrm{.}$ The right side has factor $\exp \text{\hspace{0.17em}⁣}(\frac{\eta s^2}{2}K(f,f)+\frac{\eta t^2}{2}K(g,g)+\eta st(W+K)(f,g))\mathrm{.}$ Symmetry and bilinearity of $K$ make these exponents equal.

Comparing coefficients of $s,t$ in step 3.1 first proves the identity for powers of linear functionals. Replacing $f$ by ${\sum }_j{s}_j{f}_j$ and comparing the coefficient of $s_1\cdots s_m$, and doing the same independently for $g$, proves it for every finite product of distinct linear functionals. Each fixed coefficient is continuous in its smooth compact kernels by [F2]–[F4]; the density in [F1] therefore extends the identity to all $F,G\in {\mathcal{P}}_{\mathrm{r}\mathrm{e}\mathrm{g}}(X)$. For an input of degree $d$, the coefficient of ${\mathrm{\Gamma }}_K^r$ in ${\alpha }_K{\alpha }_{-K}$ is multiplied by ${\sum }_{j=0}^r\left(\genfrac{}{}{0px}{}{r}{j}\right)(-1{)}^{r-j}=(1-1{)}^r$, which is zero for $r>0$ and one for $r=0$. The sums are finite because ${\mathrm{\Gamma }}_K$ lowers degree by two, so ${\alpha }_K{\alpha }_{-K}=I$; the reverse composition is identical.