Two measuring scales, one Weyl product

Two symbols can have different derivative scales. Their Weyl product couples those scales through the symplectic form. The useful parameter measures the coupling; it need not control either scale separately. We identify the exact compatibility conditions, derive the quantization formula, and prove product and remainder estimates.

Localizing symbols with moving metrics supplies symbol spaces, partitions and bounded compactly supported approximation. Quadratic Fourier multipliers at a moving scale supplies the Fourier estimates with convention $D=-i\partial$. We also use finite-dimensional linear algebra; completeness, compact smooth approximation and integration in $L^2$; and the tempered Schwartz kernel theorem: every continuous linear map $\mathcal S(V)\to\mathcal S'(V)$ has a unique kernel in $\mathcal S'(V\times V)$. The kernel theorem is used only for the converse description of all Schwartz-to-distribution operators. Sections 4.1–4.5 below prove that theorem with the original Schwartz seminorms and continuity into the strong dual. The operators and kernel identities needed for the product are constructed directly. Basic references are [Lerner, Metrics in phase space].

Throughout, $V$ is a real vector space of dimension $n\geq1$, $W=V\oplus V^*$, and $$\sigma((x,\xi),(y,\eta))=\langle\xi,y\rangle-\langle x,\eta\rangle. \tag{W1}$$ Measures on $V$ and $V^*$ are dual for Fourier inversion with coefficient $(2\pi)^{-n}$. Multiplying one measure by a positive constant divides the other by that constant; the phase-space measure and all quantizations below are unchanged.

1. Quadratic forms and the cross parameter

For a positive-definite quadratic form $q$ on $W$, define $$q^\sigma(T)=\sup_{S\ne0}\frac{|\sigma(T,S)|^2}{q(S)}. \tag{W2}$$ This is positive definite. Ordinary quadratic duality, composed with the invertible linear map induced by $\sigma$, gives $$(q^\sigma)^\sigma=q,\qquad (c q)^\sigma=c^{-1}q^\sigma, \qquad q\leq C r\ \Longleftrightarrow\ r^\sigma\leq Cq^\sigma. \tag{W3}$$ For example, write $q(T)=T^tQT$ in coordinates and let $J$ represent $\sigma$. Then $q^\sigma(T)=T^tJQ^{-1}J^tT$; substitution proves the first identity, and inversion of positive matrices proves the order reversal. Alternatively, all three follow by mapping the unit ellipsoid to its polar ellipsoid and using that taking the polar twice returns the original ellipsoid.

We will need an exact formula for the dual of a sum. If $F_1,F_2$ are positive quadratic forms on a finite-dimensional space and primes denote ordinary dual forms, then $$(F_1+F_2)'(\zeta) =\min_{\zeta_1+\zeta_2=\zeta} \big(F_1'(\zeta_1)+F_2'(\zeta_2)\big). \tag{W4}$$ To prove it, represent the forms by positive matrices $A_1,A_2$. Put $v=(A_1+A_2)^{-1}\zeta$ and $\zeta_j=A_jv$. Every other decomposition is $(\zeta_1+e,\zeta_2-e)$. Expanding its objective gives the value $v^t(A_1+A_2)v$, plus $e^t(A_1^{-1}+A_2^{-1})e$; the mixed terms cancel. The latter quantity is nonnegative and vanishes only at $e=0$. This proves both the minimum and its value. In particular, if $g=(g_1+g_2)/2$, $$g^\sigma(T)=2\min_{T_1+T_2=T} \big(g_1^\sigma(T_1)+g_2^\sigma(T_2)\big). \tag{W5}$$ The factor two comes from the factor one half in the mean; it is not optional.

At one point of phase space, set $$h_j^2=\sup_{T\ne0}\frac{g_j(T)}{g_j^\sigma(T)},\qquad H^2=\sup_{T\ne0}\frac{g_1(T)}{g_2^\sigma(T)} =\sup_{T\ne0}\frac{g_2(T)}{g_1^\sigma(T)}. \tag{W6}$$ The two expressions for $H$ are equal by (W3): each is the least constant in one of the equivalent inequalities $g_1\leq H^2g_2^\sigma$, $g_2\leq H^2g_1^\sigma$. These are also equivalent to $$|\sigma(T,S)|^2\leq H^2g_1^\sigma(T)g_2^\sigma(S). \tag{W7}$$ Indeed, divide by $g_2^\sigma(S)$, take the supremum over $S$, and use (W3).

For the mean metric, write $h_g^2=\sup g/g^\sigma$. Then $$\max(h_1^2,h_2^2,H^2)\leq4h_g^2 \leq h_1^2+h_2^2+2H^2. \tag{W8}$$ For the first inequality, $g_j\leq2g$ and $g^\sigma\leq2g_k^\sigma$ show that $g_j(T)/g_k^\sigma(T)\leq4g(T)/g^\sigma(T)$, for both choices of $j,k$. For the second, $2g=g_j+g_k\leq(h_j^2+H^2)g_j^\sigma$. Dualizing gives $2g_j\leq(h_j^2+H^2)g^\sigma$. Add the two inequalities. No uncertainty inequality has been assumed in this argument.

2. Transport between compatible metrics

Let $g_1,g_2$ be slowly varying metrics on $W$, in the sense of Localizing symbols with moving metrics. Write $q_j(X)=g_{j,X}^\sigma$. Each metric is assumed symplectically temperate, meaning that for fixed constants $C,N$, uniformly in $X,Y,T$, $$q_j(X)(T)\leq Cq_j(Y)(T) \big(1+q_j(Y)(X-Y)\big)^N. \tag{W9}$$ By (W3), this is equivalent to the primal inequality $g_{j,Y}\leq Cg_{j,X}(1+q_j(Y)(X-Y))^N$. Constants can be enlarged so that a common $C\geq1$, $N\geq0$ works for both metrics. The bases of the quadratic distances matter.

The cross tests are $$\begin{split} q_1(X)(T)&\leq Cq_1(Y)(T) \big(1+q_2(X)(Y-X)\big)^N,\\ q_2(X)(T)&\leq Cq_2(Y)(T) \big(1+q_1(X)(Y-X)\big)^N. \end{split} \tag{W10}$$ We call the pair compatible when these hold. They require more than temperateness of each metric separately.

Lemma 2.1 (metric transport). Under (W9)–(W10), there are common constants $C,L$ such that, for all $j,k\in\{1,2\}$, $$q_j(X)\leq Cq_j(Y)\big(1+q_k(Y)(X-Y)\big)^L, \quad g_{j,Y}\leq Cg_{j,X}\big(1+q_k(Y)(X-Y)\big)^L. \tag{W11}$$ Moreover, with $g=(g_1+g_2)/2$, there are $C',L'$ such that $$g_{j,Y}\leq C'g_{j,X} \big(1+g_Y^\sigma(X-Y)\big)^{L'},\qquad j=1,2. \tag{W12}$$ Consequently $g$ is symplectically temperate.

Proof. For $j=k$, the first estimate in (W11) is (W9). For $j\ne k$, (W9) gives $$1+q_k(X)(X-Y)\leq C_0\big(1+q_k(Y)(X-Y)\big)^{N+1}.$$ Insert this into (W10). Dualizing gives the second estimate in (W11).

Fix $X,Y$, choose any intermediate point $Z$, and put $$R=1+q_1(Y)(X-Z)+q_2(Y)(Z-Y).$$ Use (W11) first on the segment from $Z$ to $Y$, with $k=2$. It gives $$g_{j,Y}\leq Cg_{j,Z}R^L, \qquad q_1(Z)(X-Z)\leq Cq_1(Y)(X-Z)R^L\leq CR^{L+1}.$$ Use (W11) again on the segment from $X$ to $Z$, now with $k=1$. Thus $$g_{j,Z}\leq Cg_{j,X}(1+q_1(Z)(X-Z))^L \leq C_1g_{j,X}R^{L(L+1)}.$$ Multiplication gives an exponent $L^2+2L$. Minimize $R$ using (W5); its minimum is $1+g_Y^\sigma(X-Y)/2$. This proves (W12). Adding the two primal inequalities proves temperateness of the mean. It is also slowly varying: if $g_X(Y-X)$ is sufficiently small, then $g_{j,X}(Y-X)\leq2g_X(Y-X)$ is small for both metrics, and their slow-variation comparisons can be added. ∎

A positive weight $m$ is symplectically temperate for $g$ if it is $g$-continuous locally and $$m(Y)\leq Cm(X)\big(1+g_Y^\sigma(X-Y)\big)^N. \tag{W13}$$ Its reciprocal has the same property, with possibly larger exponent. In fact, apply (W13) with $X,Y$ exchanged and use (W9) to replace the resulting distance based at $X$ by a power of the distance based at $Y$. Local comparability also reverses. Products and positive real powers are handled in the same way.

Suppose $m_j$ is initially temperate for its own $g_j$. The additional weight tests are $$m_1(Y)\leq Cm_1(X)(1+q_2(X)(X-Y))^N, \quad m_2(Y)\leq Cm_2(X)(1+q_1(X)(X-Y))^N. \tag{W14}$$ For a compatible pair, these are equivalent to both weights being temperate for the mean $g$. Here is the full reduction. Own temperateness and (W14), with the distance-base conversion used above, give $m_j(Y)\leq Cm_j(X)(1+q_k(Y)(X-Y))^L$ for either $k$. Repeat the two-segment proof of (W12), replacing its primal metric factor by $m_j$; the estimate for $q_1(Z)$ is unchanged. Minimization proves (W13) for the mean. Local $g$-continuity follows from $g_j$-continuity and $g_j\leq2g$.

Conversely, $g_Y^\sigma\leq2q_k(Y)$, followed by the distance-base conversion for $g_k$, turns mean temperateness into (W14). If only mean temperateness and $g_j$-continuity were given at the outset, the same inequality with $k=j$ also proves own temperateness. This explains the exact weight hypotheses used later.

3. What the diagonal test requires

On $W\times W$, use the product metric and weight $$G_{(Y,Z)}(T,S)=g_{1,Y}(T)+g_{2,Z}(S),\qquad M(Y,Z)=m_1(Y)m_2(Z). \tag{W15}$$ The product metric is slowly varying and the product weight is locally $G$-continuous, directly from the separate local comparisons.

Use dual coordinates $(p,q;r,s)$ and the auxiliary quadratic phase $A_0(p,q;r,s)=2(q\cdot r-p\cdot s)$. Its symmetric map in the convention of Quadratic Fourier multipliers at a moving scale sends $(p,q;r,s)$ to $(-s,r;q,-p)$. Substitution in that lesson’s definition of the phase-dual form gives $$G_{(Y,Z)}^{A_0}(T,S)=q_2(Z)(T)+q_1(Y)(S). \tag{W16}$$ Consequently $G_{(X,X)}\leq H(X)^2G_{(X,X)}^{A_0}$, and $H(X)^2$ is the least possible constant. Notice the interchange of the two metrics in (W16).

Theorem 3.1 (diagonal criterion). The product metric is uniformly $A_0$-temperate at the points $(X,X)$ if and only if (W10) holds. If the individual weights are temperate for their own metrics, the product weight is uniformly $A_0,G$-temperate there if and only if (W14) holds.

Proof of necessity. In dual form, metric temperateness on the diagonal says $$q_1(X)(T)+q_2(X)(S) \leq C\big(q_1(Y)(T)+q_2(Z)(S)\big)\mathcal D^L, \quad \mathcal D=1+q_2(Z)(X-Y)+q_1(Y)(X-Z). \tag{W17}$$ Set $Z=X,S=0$ to obtain the first cross test. Set $Y=X,T=0$ for the second. The analogous product-weight inequality is $m_1(Y)m_2(Z)\leq Cm_1(X)m_2(X)\mathcal D^L$. The same substitutions and cancellation of a positive factor give (W14).

Proof of sufficiency. We supply the distance comparison that is needed for this implication. For any single temperate metric $g$, write $q(U)=g_U^\sigma$, and let $$\mathcal E=1+q(Y)(X-Z)+q(Z)(X-Y),\qquad P=Y+Z-X.$$ Because $P-Z=Y-X$ and $P-Y=Z-X$, temperateness gives $$q(P)(X-Y)\leq C\mathcal E^{L+1},\qquad q(P)(X-Z)\leq C\mathcal E^{L+1}.$$ For example, the first is the comparison of $q(P)$ with $q(Z)$, whose controlling distance $P-Z$ is exactly $-(X-Y)$. Compare $q(Y)$ with $q(P)$; the controlling distance $Y-P=X-Z$ is bounded by the second estimate. Do the analogous comparison for $q(Z)$. We obtain $$q(Y)(X-Y)+q(Z)(X-Z)\leq C'\mathcal E^{(L+1)^2}. \tag{W18}$$

Now take the mean metric from Section 2. Its dual is at most twice each $q_j$, so $\mathcal E\leq2\mathcal D$. Apply (W12) to $g_{1,Y}/g_{1,X}$ and $g_{2,Z}/g_{2,X}$, and then (W18). This yields $$g_{1,Y}(T)+g_{2,Z}(S) \leq C''\big(g_{1,X}(T)+g_{2,X}(S)\big)\mathcal D^{L''}.$$ It is precisely the primal phase-temperateness condition from Section 5 of Quadratic Fourier multipliers at a moving scale. Dualization gives (W17). The weights are temperate for the mean by the preceding weight equivalence, so applying (W13) and (W18) also proves the product-weight condition. ∎

Taking $g_1=g_2=g$ makes the cross tests consequences of ordinary symplectic temperateness and its distance-base conversion. Thus one temperate metric and any two temperate weights always satisfy the diagonal conditions, with parameter $H=h_g$. This includes metrics and weights with no initial ordinary continuity.

4. Kernels and the Weyl normalization

For $a\in\mathcal S(W)$ and $\tau\in\mathbb R$, define $$\operatorname{Op}_\tau(a)u(x) =(2\pi)^{-n}\iint e^{i\langle x-y,\xi\rangle} a((1-\tau)x+\tau y,\xi)u(y)\,dy\,d\xi. \tag{W19}$$ The choices $\tau=0,1,1/2$ are left, right and Weyl quantization. We write $a^w=\operatorname{Op}_{1/2}(a)$.

These definitions extend to every tempered distribution $a$. For example, put $$K_a(z+t/2,z-t/2) =(2\pi)^{-n}\int e^{i\langle t,\xi\rangle}a(z,\xi)\,d\xi, \qquad a(z,\xi)=\int e^{-i\langle t,\xi\rangle} K_a(z+t/2,z-t/2)\,dt. \tag{W20}$$ Both equalities are identities of tempered distributions: partial Fourier transformation and the invertible linear coordinate map $(z,t)\mapsto(z+t/2,z-t/2)$ are continuous isomorphisms of Schwartz space and its dual. The absolute determinant of this coordinate map is one.

For $u,v\in\mathcal S(V)$, define $\langle a^wu,v\rangle$ by applying $K_a$ to $u(y)\overline{v(x)}$. This is a continuous map $\mathcal S\to\mathcal S'$. Indeed, each Schwartz seminorm of the test tensor is bounded by a product of finitely many seminorms of $u,v$; a tempered distribution is bounded by a finite sum of such seminorms. The same construction works for every $\tau$. It also gives the usual weak integral interpretation of (W19). The kernel theorem contract supplies the converse: all continuous maps $\mathcal S\to\mathcal S'$ are obtained uniquely in this way.

For a finite polynomial in frequency, $a(x,\xi)=\sum_\alpha a_\alpha(x)\xi^\alpha$, left and right quantization put the coefficients on the corresponding sides: $$\operatorname{Op}_0(a)u=\sum_\alpha a_\alpha D^\alpha u, \qquad \operatorname{Op}_1(a)u=\sum_\alpha D^\alpha(a_\alpha u).$$ This remains valid for coefficients $a_\alpha\in\mathcal S'(V)$ and $u\in\mathcal S(V)$. Indeed, the inverse Fourier transform of $\xi^\alpha$ is $D_t^\alpha\delta(t)$. In the left kernel the coefficient is evaluated at $x$, so pairing this delta derivative against the input differentiates $u$; in the right kernel it is evaluated at $y$, so the output derivative acts on the entire product $a_\alpha u$. These kernels are defined by tensor products followed by invertible linear coordinate changes, and multiplication of a tempered coefficient by the smooth Schwartz input is well defined. No product of two arbitrary distributions is used.

Taking the complex conjugate and exchanging $x,y$ in (W20) proves $$(a^w)^*=(\overline a)^w. \tag{W21}$$ Here the adjoint is the distributional sesquilinear adjoint on Schwartz functions. A real symbol therefore gives a symmetric Schwartz-domain operator whenever it has values in $L^2$. Equation (W21) alone makes no assertion of selfadjoint closure for a general real symbol.

4.1. Completeness with the original Schwartz seminorms

We prove the kernel theorem used above, including continuity into the strong dual. The argument also permits different finite input and output dimensions. Work first on $\mathbb R^p$ and $\mathbb R^q$, with original positive constant densities $d\mu_x=c_x\,dx$, $d\mu_y=c_y\,dy$. The constants $c_x,c_y>0$ remain in every coefficient integral. Choosing a basis in the original $V$ gives the case $p=q=n$ with its actual density, not a replacement measure. The distributions are complex-linear; use the bilinear pairing $B(f,g)=(Af)(g)$. In the sesquilinear notation above the same value is $\langle Af,v\rangle=B(f,\overline v)$.

Retain every monomial seminorm $$p_{\alpha,\beta}(f)=\sup_x|x^\alpha\partial^\beta f(x)|, \qquad P_N(f)=\max_{|\alpha|\leq N,\ |\beta|\leq N}p_{\alpha,\beta}(f), \quad N=0,1,2,\ldots . \tag{WK1}$$ The finite maxima $P_N$ express the same topology because each original seminorm occurs in some maximum, and every maximum involves only finitely many original seminorms. The metric and topological arguments in Section 14.1 of Banach estimates, quotient spaces and compact parameter arguments apply to this countable separating family.

Here is completeness itself. If $f_j$ is Cauchy for every original seminorm, then $\partial^\beta f_j$ converges uniformly on all of $\mathbb R^p$ to a continuous function $g_\beta$. Scalar completeness and the uniform Cauchy estimate give that limit. The fundamental theorem on a coordinate segment gives $$\partial^\beta f_j(x+te_i)-\partial^\beta f_j(x) =\int_0^t\partial^{\beta+e_i}f_j(x+se_i)\,ds. \tag{WK2}$$ Uniform convergence passes this identity to $g_\beta,g_{\beta+e_i}$. Differentiating the integral proves that $g_0$ is smooth with derivatives $g_\beta$, by induction on the full multi-index. For each $\alpha,\beta$, the weighted functions $x^\alpha\partial^\beta f_j$ are also uniformly Cauchy. Their pointwise limit is $x^\alpha g_\beta$, so they converge uniformly to it. That limit is bounded, and $f_j\to g_0$ in that original seminorm. Thus $g_0\in\mathcal S$ and the space is Fréchet. The proof includes all polynomial weights, not only smooth convergence on compact sets. It applies in dimension $p+q$ as well.

We will estimate some sums with auxiliary polynomial weights, while keeping their comparison with (WK1) explicit. For an integer $S\geq0$, the complete multinomial identity yields $$(1+|x|)^S\leq(1+\textstyle\sum_i|x_i|)^S =\sum_{h+|\alpha|=S}\frac{S!}{h!\alpha!}|x^\alpha|. \tag{WK3}$$ Consequently a supremum weighted by $(1+|x|)^S$ and involving derivatives through order $R$ is bounded by the entire finite sum of the corresponding original $p_{\alpha,\beta}$, with coefficients in (WK3). Conversely $p_{\alpha,\beta}$ is bounded by that weighted supremum with $S=|\alpha|$, $R=|\beta|$.

On the product use $$Q_{S,R}(\Phi)= \max_{|\gamma|+|\delta|\leq R} \sup_{x,y}(1+|x|)^S(1+|y|)^S |\partial_x^\gamma\partial_y^\delta\Phi(y,x)|. \tag{WK4}$$ It is bounded by the full product of the two multinomial sums (WK3): explicitly, sum over $h+|\alpha|=S$, $j+|\beta|=S$ with coefficient $(S!)^2/(h!\alpha!j!\beta!)$, multiplying $\max_{|\gamma|+|\delta|\leq R}\sup|x^\alpha y^\beta\partial_x^\gamma\partial_y^\delta\Phi|$. This proves continuity of (WK4) in the original product Schwartz topology, and records all weights and coefficients.

4.2. The global bilinear estimate

For separately continuous bilinear $B:\mathcal S(\mathbb R^p)\times\mathcal S(\mathbb R^q)\to\mathbb C$, there are $N,M\geq0$ and $C>0$ such that $$|B(f,g)|\leq C P_N(f)P_M(g). \tag{WK5}$$ To prove it, for $m\geq1,M\geq0$ form the closed sets $$E_{m,M}=\{f:|B(f,g)|\leq mP_M(g)\text{ for every }g\}. \tag{WK6}$$ Separate continuity makes them closed, and continuity of each $B(f,\cdot)$ makes their countable union the whole first Schwartz space. Completeness from Section 4.1 and the complete-metric Baire theorem in Section 6 of the Banach foundation lesson give one set with interior. Take $f_0$ in that interior and $r>0,N$ with $f_0+h$ in that set whenever $P_N(h)<r$. Subtract its bounds for $f_0+h,f_0$ to obtain $2mP_M(g)$. Rescale $h$ by $r/(2P_N(h))$ to give (WK5) with $C=4m/r$. The zero-seminorm case gives zero by arbitrary rescaling; here $P_N$ already separates points. Thus (WK5) is a global estimate, with fixed finite orders for all Schwartz inputs.

4.3. A lattice expansion with all weights and periods

Choose a nonnegative smooth bump $\theta$ supported in $(-1,1)$, positive on $[-1/2,1/2]$. Put $$h(t)=\sum_{\nu\in\mathbb Z}\theta(t-\nu),\qquad \chi_1(t)=\theta(t)/h(t). \tag{WK7}$$ The sum is locally finite and is smooth, positive and one-periodic. Its minimum on $[0,1]$ is positive, so the quotient is smooth and compactly supported. Periodicity gives $\sum_\nu\chi_1(t-\nu)=1$. Define $\chi(x)=\prod_{i=1}^p\chi_1(x_i)$, $\eta(y)=\prod_{j=1}^q\chi_1(y_j)$, and their translates $\chi_a(x)=\chi(x-a)$, $\eta_b(y)=\eta(y-b)$, for $a\in\mathbb Z^p,b\in\mathbb Z^q$. Their locally finite sums are both one. All translate derivatives have their original, translate-independent bounds.

Choose periods $\ell_i>4,s_j>4$, and the boxes $$I_0=\prod_i(-\ell_i/2,\ell_i/2),\quad J_0=\prod_j(-s_j/2,s_j/2),\quad I_a=a+I_0,\quad J_b=b+J_0,$$ $$m_I=\mu_x(I_0)=c_x\prod_i\ell_i,\qquad m_J=\mu_y(J_0)=c_y\prod_j s_j. \tag{WK8}$$ Fix $\sigma\in\mathcal D(I_0)$, $\tau\in\mathcal D(J_0)$, equal to one near the closed unit coordinate cubes. Write $\sigma_a(x)=\sigma(x-a)$, $\tau_b(y)=\tau(y-b)$, and $$e_{a,k}(x)= \exp\!\left(2\pi i\sum_i k_i(x_i-a_i+\ell_i/2)/\ell_i\right), \quad e_{b,l}(y)= \exp\!\left(2\pi i\sum_j l_j(y_j-b_j+s_j/2)/s_j\right). \tag{WK9}$$ For $\Phi\in\mathcal S(\mathbb R^q\times\mathbb R^p)$, localize by $\eta_b(y)\chi_a(x)$. Its coefficient is $$c_{b,a;l,k}(\Phi)=\frac1{m_Jm_I} \int_{J_b}\int_{I_a} \eta_b(y)\chi_a(x)\Phi(y,x)e_{b,-l}(y)e_{a,-k}(x) \,d\mu_x(x)\,d\mu_y(y). \tag{WK10}$$ The factors $c_x,c_y$ occur in both the numerator measures and the denominator volumes; this equality is also the ordinary periodic Fourier coefficient in these coordinates. It does not change the original density of any kernel pairing.

The localized test vanishes near all period-box faces. For $$L=1-\sum_i(\ell_i/(2\pi))^2\partial_{x_i}^2 -\sum_j(s_j/(2\pi))^2\partial_{y_j}^2,$$ $$L^r=\sum_{h+|\alpha|+|\beta|=r} \frac{r!(-1)^{|\alpha|+|\beta|}}{h!\alpha!\beta!} \prod_i(\ell_i/(2\pi))^{2\alpha_i} \prod_j(s_j/(2\pi))^{2\beta_j} \partial_x^{2\alpha}\partial_y^{2\beta}, \tag{WK11}$$ integration by parts replaces the integrand in (WK10) by $L^r(\eta_b\chi_a\Phi)$ and divides the coefficient by $(1+|k|^2+|l|^2)^r$. There are no face terms, and the constant summand in (WK11) is retained.

For clarity we also prove the periodic inversion used here. On a circle of length $\ell$, the nonnegative Fejér kernel is $$F_T(t)=\frac1{T+1}\left|\sum_{\nu=0}^T e^{2\pi i\nu t/\ell}\right|^2 =\sum_{|k|\leq T}\left(1-\frac{|k|}{T+1}\right)e^{2\pi ikt/\ell}. \tag{WK12}$$ Its integral divided by $\ell$ is one. At circular distance at least $\delta$, $0<\delta<\ell/2$, the geometric-sum formula bounds it by $[(T+1)\sin^2(\pi\delta/\ell)]^{-1}$; hence its mass there has that same upper bound. Product kernels over all input and output circles, with each factor $1/\ell_i$ or $1/s_j$ in convolution, have total mass one and tail mass bounded by the sum of the one-coordinate tails. Uniform continuity on the compact period box then proves uniform convergence of convolution: the difference inside small coordinate intervals is bounded by the modulus of continuity, and outside by twice the supremum times the tail mass. Apply this to every derivative of a smooth periodic function; differentiation commutes with the finite integral.

These convolutions are finite Fourier sums with coefficient weights $\prod_i(1-|k_i|/(T+1))_+\prod_j(1-|l_j|/(T+1))_+$. Formula (WK11), with arbitrarily large $r$, makes each differentiated unweighted Fourier series absolutely uniformly convergent. Indeed a dyadic lattice shell in dimension $d$ contains at most $(2^{v+2}+1)^d$ vectors; decay of exponent greater than $d$ makes its shell sum geometric. Dominated convergence of the coefficient series makes its Fejér sums tend to the unweighted sum. The convolution limit identifies that sum with the original function. The fundamental theorem on coordinate segments verifies termwise derivatives from their uniform convergence. Multiplying by $\sigma_a,\tau_b$ now gives $$\eta_b(y)\chi_a(x)\Phi(y,x)= \sum_{l,k}c_{b,a;l,k}(\Phi) \tau_b(y)e_{b,l}(y)\sigma_a(x)e_{a,k}(x) \tag{WK13}$$ with convergence in every fixed-support smooth seminorm.

We require decay in the lattice centers as well as the frequencies. On the support of $\eta_b\chi_a$ or any of its derivatives, $$1+|a|\leq(1+\sqrt p)(1+|x|),\qquad 1+|b|\leq(1+\sqrt q)(1+|y|). \tag{WK14}$$ The complete product rule in (WK11) therefore gives $$|c_{b,a;l,k}(\Phi)| \leq \frac{C_r(1+\sqrt p)^S(1+\sqrt q)^S Q_{S,2r}(\Phi)} {(1+|a|)^S(1+|b|)^S(1+|k|^2+|l|^2)^r}, \tag{WK15}$$ where the whole constant is $$\begin{aligned} C_r={}&\frac1{m_Jm_I} \sum_{h+|\alpha|+|\beta|=r} \frac{r!}{h!\alpha!\beta!} \prod_i(\ell_i/(2\pi))^{2\alpha_i} \prod_j(s_j/(2\pi))^{2\beta_j} \\ &{}\times\sum_{\gamma\leq2\alpha,\ \delta\leq2\beta} \binom{2\alpha}{\gamma}\binom{2\beta}{\delta} \int_{J_0}\int_{I_0} |\partial^\gamma\chi(x)|\,|\partial^\delta\eta(y)| \,d\mu_x(x)\,d\mu_y(y). \end{aligned} \tag{WK16}$$ All cutoff derivatives and the positive coordinate-density factors are included. The remaining derivative of $\Phi$ in each summand has order at most $2r$, as recorded in (WK4). The support bound (WK14) applies even if one of the cutoff derivatives vanishes.

For every original $P_N$, the complete product rule also gives $$P_N(\sigma_a e_{a,k}) \leq D_{\sigma,N}(1+|a|)^N(1+|k|)^N,$$ $$D_{\sigma,N}= \max_{|\alpha|,|\beta|\leq N} \prod_i\max\{1,\ell_i/2\}^{\alpha_i} \sum_{\gamma\leq\beta}\binom{\beta}{\gamma} \|\partial^\gamma\sigma\|_\infty \prod_i(2\pi/\ell_i)^{\beta_i-\gamma_i}. \tag{WK17}$$ To check it, on the translated support $|x_i|\leq|a_i|+\ell_i/2\leq\max\{1,\ell_i/2\}(1+|a|)$. The derivative of the exponential supplies exactly $\prod_i(2\pi|k_i|/\ell_i)^{\beta_i-\gamma_i}$, with zero powers equal to one. Since $|k_i|\leq1+|k|$, and both total orders are at most $N$, this proves (WK17). The analogous constant $D_{\tau,M}$, with periods $s_j$ and output dimension $q$, gives the output bound.

The whole sum of (WK13) over $a,b$ converges in the original product Schwartz topology: $$\Phi=\sum_{a,b,k,l}c_{b,a;l,k}(\Phi) \tau_b e_{b,l}\otimes\sigma_a e_{a,k}. \tag{WK18}$$ For any requested monomial/derivative seminorm choose an integer $P$ bounding its input and output orders. The seminorm of each tensor is bounded by the product of the two estimates (WK17) with order $P$. Choose $S>P+p,\ S>P+q$ and $2r>2P+p+q$ in (WK15). The sums over centers and frequencies converge by the dyadic shell count just proved. Therefore the series is absolutely Cauchy in every original seminorm. Completeness in Section 4.1 supplies a Schwartz limit. At a fixed point only finitely many centers have $\sigma_a$ or $\tau_b$ nonzero. Formula (WK13) sums those local pieces, and the sums of $\chi_a,\eta_b$ are one; hence the pointwise limit is $\Phi$. This identifies the Schwartz limit and proves that finite sums of product Schwartz functions are dense. It also proves the one-factor version of (WK18), reconstructing every $f\in\mathcal S(\mathbb R^p)$ and $g\in\mathcal S(\mathbb R^q)$ by their own localized coefficient series.

4.4. A tempered kernel for every separately continuous pairing

For $B$ with the global bound (WK5), define $$K(\Phi)=\sum_{a,b,k,l}c_{b,a;l,k}(\Phi) B(\sigma_a e_{a,k},\tau_b e_{b,l}). \tag{WK19}$$ This is an absolutely convergent scalar series and a tempered distribution. In fact (WK5), (WK15) and (WK17) give the full bound $$|K(\Phi)|\leq C C_rD_{\sigma,N}D_{\tau,M} (1+\sqrt p)^S(1+\sqrt q)^S Q_{S,2r}(\Phi)$$ $${}\times \left(\sum_a(1+|a|)^{N-S}\right) \left(\sum_b(1+|b|)^{M-S}\right) \left(\sum_{k,l} \frac{(1+|k|)^N(1+|l|)^M} {(1+|k|^2+|l|^2)^r}\right), \quad S>N+p,\quad S>M+q,\quad 2r>N+M+p+q. \tag{WK20}$$ Each sum is finite by its precise strict inequality and the dyadic shell count. Formula (WK4) bounds the entire right side by a finite sum of original product Schwartz seminorms. Linearity and continuity of (WK19) follow. No compact-support estimate is being extended without these center-weight sums.

If $\Phi(y,x)=g(y)f(x)$, Fubini in (WK10) gives the product of the one-factor coefficients, with both original volume factors retained. The one-factor reconstructions in Section 4.3 converge in the two original Schwartz topologies. Bilinear continuity (WK5), followed by the absolute convergence of (WK19), proves $$K(g\otimes f)=B(f,g). \tag{WK21}$$ Product density (WK18) proves uniqueness of this tempered distribution and independence of the auxiliary bump, periods, boxes and cutoffs. Conversely a tempered distribution $K$ has a finite-order bound by one finite maximum of original product seminorms, and the product rule for $g(y)f(x)$ gives a bound of the form (WK5) for $B(f,g)=K(g\otimes f)$. Thus this construction is exactly a bijection between tempered kernels and separately continuous bilinear Schwartz pairings.

4.5. The exact strong-dual operator theorem

The strong topology on $\mathcal S'(\mathbb R^q)$ is defined by $$T_H(u)=\sup_{g\in H}|u(g)|, \qquad H\subset\mathcal S(\mathbb R^q)\text{ bounded}. \tag{WK22}$$ Boundedness here means boundedness for every original monomial seminorm, equivalently for every $P_M$. Indeed every neighborhood contains finitely many of those seminorm constraints, and a single dilation contains the set exactly when their suprema are finite. In particular every singleton is bounded.

If $A:\mathcal S(\mathbb R^p)\to\mathcal S'(\mathbb R^q)$ is continuous into this strong dual, $B(f,g)=(Af)(g)$ is separately continuous: evaluation at a fixed $g$ is one strong-dual seminorm, and for a fixed $f$ the value $Af$ is a tempered distribution. The preceding sections give its unique kernel.

Conversely a kernel gives $Af$ by $(Af)(g)=K(g\otimes f)$. Its finite product-seminorm estimate has the form $|K(g\otimes f)|\leq C P_N(f)P_M(g)$. Thus $Af$ is a tempered distribution and, for every bounded $H$, $$T_H(Af)\leq C P_N(f)\sup_{g\in H}P_M(g). \tag{WK23}$$ This proves actual continuity into the strong dual. Both constructions are inverse on every pair of tests. The bijection holds also for unequal finite dimensions and for a separately continuous pairing before an operator topology has been imposed. We claim precisely these seminorm estimates and this bijection, without imposing an additional topology on the collection of all operators.

For the original $V$, a basis change expands every coordinate monomial and derivative by the full finite matrix product and multinomial rules; its inverse does the same. These two finite bounds give continuous inverse maps on the original Schwartz spaces. The kernel identification is therefore independent of the chosen basis. The densities in (WK8), (WK10) and (WK16) preserve the original coordinate-density constants. Taking $p=q=n$, and replacing $g$ by $\overline v$, recovers exactly the sesquilinear convention and the kernel class of Section 4. Conjugation preserves every original seminorm, since $x^\alpha$ is real and $\partial^\beta\overline v=\overline{\partial^\beta v}$; it is therefore an antilinear continuous bijection. The Fourier and coordinate isomorphisms in (W20), with their original $(2\pi)^{-n}$ factor and determinant one, then give the claimed converse Weyl-symbol classification.

Here is the full coordinate-seminorm comparison used in that basis assertion. If $x=Cz$ with an invertible real $p\times p$ matrix $C$, write $K,H$ for nonnegative integer $p\times p$ matrices, $K\mathbf1=\alpha$, $H\mathbf1=\beta$, and let $\kappa_j=\sum_iK_{ij}$, $\lambda_j=\sum_iH_{ij}$. The chain rule and both multinomial expansions give $$z^\alpha\partial_z^\beta(f(Cz))= \sum_{\substack{K\mathbf1=\alpha\\H\mathbf1=\beta}} \frac{\alpha!\beta!}{\prod_{i,j}K_{ij}!H_{ij}!} \prod_{i,j}(C^{-1})_{ij}^{K_{ij}}C_{ji}^{H_{ij}} \left.x^\kappa\partial_x^\lambda f(x)\right|_{x=Cz}. \tag{WK24}$$ Every summand, including zero coefficients, is retained. Since $z\mapsto x$ is onto, taking supremums and absolute values bounds each transformed original seminorm by the displayed finite sum of original $p_{\kappa,\lambda}$. Applying the same formula to $C^{-1}$ gives the inverse comparison. The density becomes $c_x|\det C|\,dz$; the output basis change has its own full positive absolute determinant. These factors stay in the numerator and volumes of (WK10) and (WK16), so the kernel pairing and the theorem are unchanged.

In dimension zero the space is $\mathbb C$; coordinate products and lattice lists have their empty-coordinate values, while its original density still has mass $c_x$ or $c_y$. Thus (WK8) retains that mass and (WK10) retains it in both numerator and denominator. The theorem reduces to the scalar pairing identity. If one dimension is zero, apply the same proof in the other factor. This also records that endpoint without changing any original positive-dimensional formula.

5. Affine observables and their unitary groups

For smooth symbols define $$\{a,b\}=\partial_\xi a\cdot\partial_xb- \partial_xa\cdot\partial_\xi b. \tag{W22}$$ When one factor is affine, this is also defined for distributional other factors. Directly from (W20), multiplication of the output by $x_j$ and differentiation by $D_{x_j}$ give $$x_j a^w=(x_ja+\tfrac i2\partial_{\xi_j}a)^w, \qquad D_{x_j}a^w=(\xi_ja+\tfrac1{2i}\partial_{x_j}a)^w.$$ For the first formula, write the output coordinate as $z_j+t_j/2$, and integrate the $t_j$ factor by parts in $\xi_j$. For the second, $\partial_{x_j}=\frac12\partial_{z_j}+\partial_{t_j}$ on the kernel. These operations on distributions are legitimate by duality. Linearity gives, for every affine $L$, $$L^w a^w=\big(La+\{L,a\}/(2i)\big)^w. \tag{W23}$$

Proposition 5.1 (affine unitary group). For real affine $L(x,\xi)=b\cdot x+c\cdot\xi+d$, the operator $L^w=b\cdot x+c\cdot D+d$, initially on $\mathcal S(V)$, is essentially selfadjoint. Its unitary group is $$U_tu(x)=e^{itd+it b\cdot x+it^2 b\cdot c/2}u(x+tc), \qquad U_t=(e^{itL})^w. \tag{W24}$$

Proof. We first obtain the $L^2$ Fourier transform from the Schwartz Plancherel prerequisite. The normalized transform $\mathcal F_0=(2\pi)^{-n/2}\mathcal F$ is an isometry on $\mathcal S$. If $u_j\to u$ in $L^2$, with $u_j\in\mathcal S$, then $\mathcal F_0u_j$ is Cauchy in $L^2$; completeness defines its limit, independently of the approximating sequence. Density therefore extends $\mathcal F_0$ to an isometry on all of $L^2$. The normalized inverse extends by the same argument. Their compositions are the identity on the dense Schwartz subspace, hence on $L^2$ by continuity. Thus the extension is unitary, using precisely Schwartz Plancherel and the declared density/completeness prerequisites.

Multiplication by a real linear function $\ell$ on $L^2$, with domain $\{u:\ell u\in L^2\}$, is selfadjoint. If a vector lies in the adjoint domain, testing against compactly supported smooth functions identifies its adjoint value as $\ell u$, so it belongs to the displayed domain; the reverse inclusion follows by integration. Compact cutoff followed by smooth convolution approximates every domain vector in the graph norm: on a fixed compact set $\ell$ is bounded, and the error in commuting convolution past $\ell$ is bounded by the mollifier radius times $|\nabla\ell|\|u\|_2$. Thus $C_c^\infty$, and hence $\mathcal S$, is a core. The unitary Fourier transform gives the same conclusion for $c\cdot D$, because its Fourier transform is multiplication by $c\cdot\xi$.

If $c=0$, this already proves the assertion. If $c\ne0$, choose orthonormal coordinates with $c=(\gamma,0,\ldots,0)$, $\gamma>0$, and put $$\phi(x)=\gamma^{-1} \left(b_1x_1^2/2+x_1\sum_{j>1}b_jx_j\right).$$ Multiplication by $e^{i\phi}$ is unitary and preserves Schwartz space, as does its inverse. Since $c\cdot\nabla\phi=b\cdot x$, $$e^{-i\phi}(c\cdot D)e^{i\phi}=c\cdot D+b\cdot x.$$ Unitary conjugation preserves selfadjointness and the core property. Addition of the real constant $d$ completes the closure argument.

The formula for $U_t$ in (W24) is unitary, preserves $\mathcal S$, and obeys $U_tU_s=U_{t+s}$. Its derivative on $\mathcal S$ is $iL^wU_t$. It is also the exponential of the closure just constructed: conjugate the translation group for $c\cdot D$ by $e^{i\phi}$; the identity $\phi(x+tc)-\phi(x)=t b\cdot x+t^2b\cdot c/2$ gives exactly (W24). The pure multiplication case is immediate. Strong continuity follows from continuity of translations in $L^2$, first for compactly supported smooth functions and then by density and the unitary norm bound. Finally, substituting $e^{itL}$ into the distributional kernel formula sets $y=x+tc$ and gives the same phase. ∎ Editorial domain calculation in the original coordinates. The affine proof can also retain the original vectors throughout. In the fixed original coordinates and density, let $c\ne0$ and define $$\phi_c(x)=\frac{(b\cdot x)(c\cdot x)}{|c|^2} -\frac{(b\cdot c)(c\cdot x)^2}{2|c|^4},\qquad M_cu=e^{i\phi_c}u. \tag{WA1}$$ Both terms, including the possibly zero second term, remain. Direct differentiation gives $$c\cdot\nabla\phi_c(x)= \frac{(b\cdot c)(c\cdot x)+(b\cdot x)|c|^2}{|c|^2} -\frac{(b\cdot c)\,2(c\cdot x)|c|^2}{2|c|^4}=b\cdot x. \tag{WA2}$$ The original Fourier transform and inverse, with their full $(2\pi)^n$ norm identity and $(2\pi)^{-n}$ inverse factor, define the closed selfadjoint operator $D_c=c\cdot D$ on $$D(D_c)=\{u\in L^2(V):c\cdot\partial u\in L^2(V) \text{ as a distribution}\}. \tag{WA3}$$ Indeed the actual transform sends this distributional derivative to multiplication by $ic\cdot\xi$; the multiplication-domain proof above and the full inverse transform give both inclusions of (WA3) and the selfadjoint domain. Its core is the inverse image of the multiplication core; that image lies in the original Schwartz space and is graph dense. Both Fourier norm factors remain in this comparison, as proved in the Fourier prerequisite and its complete real/complex receiving maps, Section17.5.

The map $M_c$ and its inverse are unitary for the original $L^2$ measure and continuous inverse maps of the original Schwartz space: each derivative is a finite polynomial times the same modulus-one exponential. The distributional product rule and (WA2) prove the exact domain and action $$\begin{aligned} D(\overline{L^w})&=\{u\in L^2: c\cdot Du+(b\cdot x+d)u\in L^2 \text{ as a distribution}\},\\ \overline{L^w}&=M_c^{-1}D_cM_c+dI \quad\text{on that entire domain}. \end{aligned} \tag{WA4}$$ The displayed sum is the condition; its two unbounded summands need not separately lie in $L^2$. Equivalence with $M_cu\in D(D_c)$ follows by subtracting the actual bounded term $du$ and applying both directions of the distributional product rule. Unitary conjugation and the proved original core give selfadjointness and essential selfadjointness on $\mathcal S$. No rotation or replacement vector is required. Moreover, $$\begin{aligned} \phi_c(x+tc)-\phi_c(x) &=\frac{(b\cdot x+t b\cdot c)(c\cdot x+t|c|^2) -(b\cdot x)(c\cdot x)}{|c|^2}\\ &\quad-\frac{(b\cdot c)((c\cdot x+t|c|^2)^2-(c\cdot x)^2)}{2|c|^4} =t b\cdot x+t^2 b\cdot c/2. \end{aligned} \tag{WA5}$$ Thus both ordered conjugation maps recover (W24), its original phase and translation. For $c=0$ the full domain is $\{u:(b\cdot x+d)u\in L^2\}$, by the same multiplication proof. If also $b=0$, this is all $L^2$, with the exact scalar action $dI$.