Symbols, operators and Sobolev scales

Frequency differentiation improves a symbol, while differentiation in the base variable may worsen it. The balance between those two effects controls an expansion. Boundedness requires a different argument: a decomposition into frequency bands continues to work when the two effects exactly balance. We develop these mechanisms separately, then combine them to construct inverses and act on Sobolev and Besov spaces.

We use the Fourier conventions proved in Fourier transforms, finite spectra and convex separation and applied in Quadratic Fourier multipliers at a moving scale: the forward kernel is $e^{-ix\cdot\xi}$, inverse transformation has factor $(2\pi)^{-n}$, and $D=-i\partial$. Fourier inversion and Plancherel, including their extension to $L^2$, are explicit prerequisites. The additional entry contracts are completeness of $L^2$, density of Schwartz functions in $L^2$, the usual integration and dominated-convergence theorems for Lebesgue measure, Taylor’s formula and the multivariable product rule. Distributional Fourier transformation, invertible linear pullback on tempered distributions, and the tempered Schwartz kernel theorem are the same explicit contracts as in Section 4 of Two measuring scales, one Weyl product. In particular the kernel theorem means that every continuous linear map $\mathcal S(\mathbb R^n)\to\mathcal S'(\mathbb R^n)$ has exactly one kernel in $\mathcal S'(\mathbb R^{2n})$, and conversely. Its full proof is Sections4.1–4.5 of the Weyl-product lesson, including the exact global kernel and strong-dual maps. Sections8.1–8.4 of the Fourier lesson prove partial transforms and the full original distributional and L2 extensions. The Banach and measure lesson supplies completeness and density; the metric and calculus lesson supplies the complete Taylor and product maps.

The multiplier results in Sections 7–8 of Quadratic Fourier multipliers at a moving scale and the classical metric verification in Section 8 of Two measuring scales, one Weyl product are available here. We use those results with specified metrics below. All statements are for a fixed finite dimension. Scalar symbols may be replaced by matrices between fixed finite-dimensional complex Hermitian spaces: norms replace absolute values, products keep their displayed order, and adjoints are conjugate transposes. The inverse theorem uses square matrices and a lower bound for their smallest singular value. No infinite-dimensional operator-valued extension is implicit.

1. Symbol classes and the order budget

Fix $$0<\rho\leq1,\qquad 0\leq\delta<1,\qquad m\in\mathbb R, \qquad \langle\xi\rangle=(1+|\xi|^2)^{1/2}. \tag{E1}$$ The space $S^m_{\rho,\delta}$ consists of smooth functions on $\mathbb R_x^n\times\mathbb R_\xi^n$ for which every seminorm $$p_{m,L}(a)=\max_{|\alpha|+|\beta|\leq L} \sup_{x,\xi}\langle\xi\rangle^{-m+\rho|\alpha|-\delta|\beta|} \|\partial_\xi^\alpha\partial_x^\beta a(x,\xi)\| \tag{E2}$$ is finite. Bounds hold for all base points with the same constants. Estimates only on compact sets in $x$ define a different, local topology. We write $S^m=S^m_{1,0}$, $S^{-\infty}=\bigcap_m S^m_{\rho,\delta}$, and $S^\infty_{\rho,\delta}=\bigcup_mS^m_{\rho,\delta}$. The intersection is independent of the parameters in (E1): it consists precisely of functions all of whose derivatives decrease faster than every power of $\langle\xi\rangle$, uniformly in $x$.

These are Fréchet spaces. Indeed, a Cauchy sequence in all seminorms has uniformly Cauchy derivatives on every compact set. Successive integration along coordinate segments identifies the limits as derivatives of a smooth function. Passing to the limit in each weighted uniform bound proves convergence in that seminorm and completeness. The seminorm family is countable and separates points.

Differentiation is continuous with its exact order change, $$\partial_\xi^\alpha\partial_x^\beta: S^m_{\rho,\delta}\longrightarrow S^{m-\rho|\alpha|+\delta|\beta|}_{\rho,\delta}. \tag{E3}$$ Multiplication is continuous bilinear from orders $m,m'$ to order $m+m'$. To see the finite-seminorm assertion, expand a derivative of $ab$: each term is $\binom\alpha\gamma\binom\beta\nu (\partial_\xi^\gamma\partial_x^\nu a) (\partial_\xi^{\alpha-\gamma}\partial_x^{\beta-\nu}b)$. The powers contributed by the two factors add to $m+m'-\rho|\alpha|+\delta|\beta|$. Summing the finitely many coefficients gives $p_{m+m',L}(ab)\leq C_Lp_{m,L}(a)p_{m',L}(b)$. This calculation also explains why exchanging two matrix factors is not allowed.

We will need one elementary stability property. If $a_1,\ldots,a_k\in S^0_{\rho,\delta}$ and $F$ is a smooth real-variable function on $\mathbb C^k$, then $F(a_1,\ldots,a_k)\in S^0_{\rho,\delta}$. Their ranges have compact closure in a bounded subset of $\mathbb C^k$, so every derivative of $F$ needed at a given stage is bounded there. An iterated chain-rule term is a derivative of $F$ times a product of positive-order derivatives of the $a_j$; their total frequency and base derivative counts are $|\alpha|$ and $|\beta|$. Equation (E2) follows. The same proof works when $F$ is defined on an open neighborhood of that compact closure. In particular inversion of a matrix-valued order-zero symbol whose inverse is uniformly bounded preserves order zero: differentiate $a^{-1}a=I$, obtaining $\partial a^{-1}=-a^{-1}(\partial a)a^{-1}$, and iterate with the original factor order.

2. Summing symbol expansions and preserving support

Frequency rescaling. If $a\in S^0$, then $a_\varepsilon(x,\xi)=a(x,\varepsilon\xi)$, $0\leq\varepsilon\leq1$, is a bounded family in $S^0$, and $$a_\varepsilon\longrightarrow a(x,0)\quad\hbox{in }S^t \quad(t>0). \tag{E4}$$ For $0<t\leq1$, each fixed $S^t$ seminorm of the difference is at most $C\varepsilon^t$, with a finite number of seminorms of $a$ in $C$. For derivatives with $\alpha=0$, boundedness and the mean-value formula give a bound $C\min(1,\varepsilon|\xi|)\leq C(\varepsilon|\xi|)^t$. For $q=|\alpha|\geq1$, derivatives of the limit vanish and $$\varepsilon^q\langle\xi\rangle^{q-t} \langle\varepsilon\xi\rangle^{-q}\leq\varepsilon^t, \tag{E5}$$ because $\varepsilon\langle\xi\rangle\leq\langle\varepsilon\xi\rangle$ and $q\geq t$. The estimates also prove boundedness at order zero. Values $t>1$ follow by the inclusion $S^1\subset S^t$. Convergence in $S^0$ need not hold: $a(\xi)=\xi_1/\langle\xi\rangle$ has $a(x,0)=0$ and $\sup_\xi|a(\varepsilon\xi)|=1$ for every positive $\varepsilon$.

Summation theorem. Let $a_j\in S^{m_j}_{\rho,\delta}$, $j\geq0$, with $m_j\to-\infty$, without assuming monotonicity. Put $M_k=\max_{j\geq k}m_j$. There exists $a\in S^{M_0}_{\rho,\delta}$ such that $$\operatorname{supp}a\subset\bigcup_j\operatorname{supp}a_j, \qquad a-\sum_{j<k}a_j\in S^{M_k}_{\rho,\delta}\quad(k\geq0). \tag{E6}$$ Any two such sums differ by $S^{-\infty}$. Permuting the terms does not change their sum modulo that space.

Proof. Choose a smooth function $\chi$ equal to one near zero and with compact support. For each $j$ choose $R_j\to\infty$, strictly increasing if necessary, so that $$\|(1-\chi(\xi/R_j))a_j\|_{m_j+1,L}\leq2^{-j} \quad(0\leq L\leq j). \tag{E7}$$ Here $\|\cdot\|_{m,L}$ means (E2). This choice is possible: on the support of the high-frequency cutoff, $\langle\xi\rangle^{-1}\to0$ uniformly as $R_j\to\infty$. A frequency derivative of order $\gamma$ hitting the cutoff gives $R_j^{-|\gamma|}$, supported where $|\xi|\asymp R_j$; since $\rho\leq1$, it is bounded by a constant times $\langle\xi\rangle^{-\rho|\gamma|}$. The remaining factors have their stipulated symbol bounds. For a fixed finite set of derivatives these estimates prove (E7).

Let $A_j=(1-\chi(\xi/R_j))a_j$ and $a=\sum_jA_j$. The sum is locally finite in frequency, hence smooth. For fixed $k,L$, choose $J\geq\max(k,L)$ so large that $m_j+1\leq M_k$ for all $j\geq J$. The tail $\sum_{j\geq J}A_j$ converges in the order-$M_k$, derivative-$L$ seminorm by (E7). The remaining terms in $a-\sum_{j<k}a_j$ are finitely many symbols of order at most $M_k$, together with finitely many $-\chi(\xi/R_j)a_j$. The latter are in $S^{-\infty}$, since their frequency support is bounded and all base derivatives are uniformly bounded there. This proves every differentiated estimate in (E6).

Support is also preserved at points outside the union. Near such a point only finitely many $A_j$ can be nonzero, and each of those vanishes in some neighborhood of the point; intersect those neighborhoods. Uniqueness follows by taking $k\to\infty$. For a permutation and a specified finite initial segment of it, let $M$ be the largest order among the omitted terms. Choose an original initial segment containing the specified terms and all terms of order greater than $M$. Subtracting its expansion expresses the new remainder as finitely many symbols of order at most $M$, plus a remainder of that order. This proves the permuted version of (E6). ∎

Recognition from values. Suppose $a$ is smooth, every one of its derivatives has some polynomial bound in $\langle\xi\rangle$, uniformly in $x$, and $$\left\|a-\sum_{j<k}a_j\right\|\leq C_k\langle\xi\rangle^{\mu_k}, \qquad \mu_k\to-\infty. \tag{E8}$$ Then $a$ has the full symbol expansion (E6). In particular no differentiated remainder assumptions are needed in (E8).

To prove this, subtract a sum $a_0$ constructed above. The difference $f=a-a_0$ decreases faster than every frequency power: for a prescribed power choose $k$ for which both $\mu_k$ and $M_k$ are sufficiently negative. Every derivative of $f$ still has a polynomial bound. If $\partial_j^2 f$ has growth exponent $q$, the one-dimensional Taylor formula in either a base or a frequency coordinate gives, for $0<h\leq1$, $$|\partial_j f(x,\xi)| \leq h^{-1}|f((x,\xi)+he_j)-f(x,\xi)| +C h\langle\xi\rangle^q. \tag{E9}$$ The weight at the shifted point is comparable to the original weight. Given an arbitrary desired decay power $N$, take $h=\langle\xi\rangle^{-L}$ with $L>q+N$, then use the already known decay of $f$ of degree greater than $L+N$. Both terms in (E9) are $O(\langle\xi\rangle^{-N})$. Thus every first derivative decreases rapidly. Reapply the same argument to each derivative already treated; the necessary second derivative has a polynomial bound by hypothesis. Induction proves rapid decrease of every derivative, so $f\in S^{-\infty}$. This proof and the summation construction work throughout (E1), including $\delta>\rho$.

Complex homogeneous expansions. For $z\in\mathbb C$ and $h=1/q$, with $q$ a positive integer, a polyhomogeneous symbol of degree $z$ and step $h$ is a symbol $a\in S^{\operatorname{Re}z}$ satisfying $a\sim\sum_{j\geq0}a_j$ in the sense of (E6), where $a_j$ is homogeneous of degree $z-jh$ outside the unit frequency ball. The convention is $t^z=e^{z\log t}$ for $t>0$. Smooth angular coefficients whose base and angular derivatives are uniformly bounded have the required symbol estimates: polar differentiation lowers the radial degree by one for each frequency derivative. Multiplying each homogeneous term by a fixed cutoff near zero makes it globally smooth; such choices differ in $S^{-\infty}$. The summation theorem then constructs the symbol. Without the uniform angular/base bounds, a homogeneous formula alone would give only a local-in-base symbol. Integer frequency differentiations change the degree by an integer, which is an integral multiple of the stipulated step $h$.

3. Quantization, kernels and signs

For every symbol in (E1), left quantization is $$\operatorname{Op}(a)u(x)=(2\pi)^{-n} \int e^{ix\cdot\xi}a(x,\xi)\widehat u(\xi)\,d\xi, \qquad u\in\mathcal S. \tag{E10}$$ This integral and all its differentiated versions converge absolutely. The map $(a,u)\mapsto\operatorname{Op}(a)u$ is continuous bilinear from $S^m_{\rho,\delta}\times\mathcal S$ into $\mathcal S$, throughout (E1). Indeed, output derivatives create finitely many powers of $\xi$ and base derivatives of $a$. To estimate $x^\gamma\partial_x^\beta\operatorname{Op}(a)u$, move $|\gamma|$ frequency derivatives off the exponential by integration by parts. Each resulting integrand is a derivative of $a$, a polynomial, and a derivative of $\widehat u$. Its absolute integral is bounded by one sufficiently large Schwartz seminorm of $u$ times finitely many seminorms (E2), after choosing the remaining weight less than $-n$. Boundary terms vanish by the same rapid decrease. This gives every Schwartz seminorm and the asserted joint continuity.

Differentiation and one integration by parts give $$[\operatorname{Op}(a),D_j]=i\operatorname{Op}(\partial_{x_j}a), \qquad [\operatorname{Op}(a),x_j]=-i\operatorname{Op}(\partial_{\xi_j}a). \tag{E11}$$ If $a(x,\xi)=\sum_\alpha a_\alpha(x)\xi^\alpha$, then $\operatorname{Op}(a)=\sum_\alpha a_\alpha(x)D^\alpha$: coefficients stand on the left. Such a polynomial belongs to a global symbol class only when its coefficients have the relevant uniform derivative bounds.

There is also an exact definition for every $a\in\mathcal S'(\mathbb R^{2n})$. Take the partial inverse Fourier transform in $\xi$, call its second variable $z$, and use the invertible linear substitution $z=x-y$. The result is a tempered kernel $K_a$. Conversely, $$K_a(x,y)=(2\pi)^{-n}\int e^{i(x-y)\cdot\xi}a(x,\xi)\,d\xi, \quad a(x,\xi)=\int e^{-iz\cdot\xi}K_a(x,x-z)\,dz \tag{E12}$$ are inverse operations on tempered distributions. The displayed integrals mean Fourier transforms there, not potentially divergent ordinary integrals. Each step is a continuous linear isomorphism, so the symbol–kernel correspondence is bijective and continuous in both directions. Combined with the explicit tempered kernel theorem contract, it identifies arbitrary continuous $\mathcal S\to\mathcal S'$ maps with tempered symbol-distributions. For smooth symbols (E12) agrees with (E10), by testing against two Schwartz functions and Fourier inversion.

Use the sesquilinear pairing $(u,v)=\int u\overline v$, linear in $u$. Transposing and conjugating the kernel sends $K_a(x,y)$ to $K_a(y,x)^*$. Its left symbol is $$a^\dagger=Q_1(a^*),\qquad Q_t=\exp\bigl(it\langle D_x,D_\xi\rangle\bigr). \tag{E13}$$ Here $Q_t$ is the Fourier multiplier $e^{itp\cdot q}$ in the variables dual to $(x,\xi)$. It preserves $\mathcal S$ and $\mathcal S'$ because all derivatives of that multiplier grow at most polynomially. To check the sign directly, apply (E12) first to $a(x,\xi)=e^{i(p\cdot x+q\cdot\xi)}$. Its operator is multiplication by $e^{ip\cdot x}$ followed by translation $u(x)\mapsto u(x+q)$. Its adjoint is therefore $v(x)\mapsto e^{-ip\cdot(x-q)}v(x-q)$, whose symbol is $e^{ip\cdot q}e^{-i(p\cdot x+q\cdot\xi)}$. This is exactly (E13). Integrating this identity against Schwartz Fourier transforms proves it for Schwartz symbols, and continuity of the kernel and Fourier operations proves it for all tempered symbols. Thus $(\operatorname{Op}(a)u,v)=(u,\operatorname{Op}(a^\dagger)v)$ whenever $u,v\in\mathcal S$, with distributional pairing when needed.

4. Gaussian estimates for the product

Assume now $\delta\leq\rho$. Set $\kappa=\rho-\delta$. Section 8 of Two measuring scales, one Weyl product applies to $$g_{(x,\xi)}(y,\eta)=\langle\xi\rangle^{2\delta}|y|^2 +\langle\xi\rangle^{-2\rho}|\eta|^2, \qquad m_a=\langle\xi\rangle^m. \tag{E14}$$ Its slow variation and temperateness hold uniformly, and its symplectic parameter is $\langle\xi\rangle^{-\kappa}$. Consequently, for every fixed real $t$, $$Q_t:S^m_{\rho,\delta}\longrightarrow S^m_{\rho,\delta}, \quad Q_ta-\sum_{|\alpha|<N}\frac{t^{|\alpha|}}{\alpha!} \partial_\xi^\alpha D_x^\alpha a \in S^{m-N\kappa}_{\rho,\delta}. \tag{E15}$$ For each output seminorm only finitely many input seminorms are used. These maps preserve convergence in the local smooth topology on bounded source sets. The phase convention matters: for $A(p,q)=t p\cdot q$, the symmetric representative is $t/2$ times the swap map, so the Gauss parameter is $|t|\langle\xi\rangle^{-\kappa}/2$. The finite-bound theorem applies to the original metric itself; the complete receiving calculation below retains every parameter-dependent factor. This proves (E15) directly from Section 8 of Quadratic Fourier multipliers at a moving scale with precisely the hypotheses checked in Section 8 of Two measuring scales, one Weyl product.

The unchanged metric and its full remainder constants.

Retain E14, E1’s parameter range and the original weight $m_a(X)=\langle\xi\rangle^m$. Assume $\delta\leq\rho$ and set $\kappa=\rho-\delta\geq0$, $X=(x,\xi)$. The original metric is

$$g_X(y,\eta)=\langle\xi\rangle^{2\delta}|y|^2+ \langle\xi\rangle^{-2\rho}|\eta|^2,\qquad m_a(X)=\langle\xi\rangle^m. \tag{EC5}$$

For $n\geq1$ and $t\ne0$, the exact phase $A_t(p,q)=tp\cdot q$ has the symmetric map

$$B_t=\frac t2\begin{pmatrix}0&I_n\\I_n&0\end{pmatrix},\qquad g_X(B_t(p,q))=\frac{t^2}{4} \bigl(\langle\xi\rangle^{2\delta}|q|^2+ \langle\xi\rangle^{-2\rho}|p|^2\bigr). \tag{EC6}$$

Taking the dual of this full positive quadratic form gives, on every original direction,

$$g_X^{A_t}(y,\eta)=\frac4{t^2} \bigl(\langle\xi\rangle^{2\rho}|y|^2+ \langle\xi\rangle^{-2\delta}|\eta|^2\bigr) =\frac4{t^2}g_X^\sigma(y,\eta),\qquad h_{g,A_t}(X)=\frac{|t|}{2}\langle\xi\rangle^{-\kappa} \leq h_*=\frac{|t|}{2}. \tag{EC7}$$

Indeed the ratio of each original metric coefficient to its phase-dual coefficient is $t^2\langle\xi\rangle^{-2\kappa}/4$. Its positive square root is the exact supremum defining $h$. No direction is lost. The phase-dual form is finite on the entire original space, since $B_t$ is invertible.

For the original temperateness and weight inequalities, retain the exact comparison

$$1+g_Y^\sigma(X-Y) =1+\frac{t^2}{4}g_Y^{A_t}(X-Y) \leq\max(1,t^2/4)\bigl(1+g_Y^{A_t}(X-Y)\bigr). \tag{EC8}$$

Thus every occurrence of an original metric constant $C_g$ with distance exponent $L_g$ becomes $C_g\max(1,t^2/4)^{L_g}$, and each original weight constant $C_m$ with its own exponent $L_m$ becomes $C_m\max(1,t^2/4)^{L_m}$. Local continuity and slow variation remain those of the unchanged $g$ and unchanged weight. The original Gauss finite-bound theorem G24–G26 applies with $h_*=|t|/2$. Its packing step is in dimension $2n$ and retains the full factor $(1+|t|/2)^{2n}$, in addition to the original structural constants and their displayed changes. That factor alone does not describe all dependence on $t$ of the theorem’s constant.

For $N,l\geq0$ and $T_j=(y_j,\eta_j)$, its exact receiving estimate is

$$\begin{split} |\partial_{T_1}\cdots\partial_{T_l}R_{N,t}a(X)| &\leq C_{N,l,t}\langle\xi\rangle^m \left(\frac{|t|}{2}\langle\xi\rangle^{-\kappa}\right)^N\\ &\quad{}\times \prod_{j=1}^{l} \left(\langle\xi\rangle^{2\delta}|y_j|^2+ \langle\xi\rangle^{-2\rho}|\eta_j|^2\right)^{1/2} p_{\leq J_{N,l,t}}(a;m_a,g),\\ R_{N,t}a&=Q_ta-\sum_{j<N}\frac{(iA_t(D))^j}{j!}a . \end{split} \tag{EC9}$$

The constants and finite derivative budget carry the actual unchanged local metric constants, the full EC8 temperateness/weight constants and the finite $h_*$ counting constants. For coordinate derivatives $\alpha$ in $\xi$ and $\beta$ in $x$, choose the corresponding coordinate directions in EC9. Their product is $\langle\xi\rangle^{-\rho|\alpha|+\delta|\beta|}$. Hence

$$|\partial_\xi^\alpha\partial_x^\beta R_{N,t}a(X)| \leq C_{N,\alpha,\beta,t} (|t|/2)^N \langle\xi\rangle^{m-N\kappa-\rho|\alpha|+\delta|\beta|} p_{\leq J_{N,\alpha,\beta,t}}(a;m_a,g). \tag{EC10}$$

The source metric seminorms are exactly controlled by finitely many E2 coordinate seminorms: expand each directional derivative into its $2n$ coordinate components and use $|y_k|\leq\langle\xi\rangle^{-\delta}g_X(T)^{1/2}$ and $|\eta_k|\leq\langle\xi\rangle^\rho g_X(T)^{1/2}$. Each ordered product expansion has finitely many terms, at most $(2n)^l$ at order $l$, and all factors remain. Conversely insert the coordinate directions to bound each coordinate seminorm by its metric seminorm. Thus EC10 proves the original E15 class inclusion with finite E2 source-seminorm control, retaining the full $(|t|/2)^N$ rather than replacing its value by one.

The Taylor coefficients are the original ones:

$$\begin{split} \sum_{j<N}\frac{(iA_t(D))^j}{j!}a &=\sum_{j<N}\sum_{|\alpha|=j} \frac{i^j t^j}{\alpha!}D_x^\alpha D_\xi^\alpha a\\ &=\sum_{j<N}\sum_{|\alpha|=j} \frac{i^j(-i)^jt^j}{\alpha!} \partial_\xi^\alpha D_x^\alpha a =\sum_{|\alpha|<N}\frac{t^{|\alpha|}}{\alpha!} \partial_\xi^\alpha D_x^\alpha a . \end{split} \tag{EC11}$$

For $N=0$ the sum is empty and $R_{0,t}=Q_t$. The finite-bound theorem also gives bounded-set local smooth convergence, continuity and uniqueness on the original symbol class. When $t=0$, $Q_0=I$: for $N\geq1$ the term of order zero is $a$ and all positive-order terms vanish, so the remainder is exactly zero; for $N=0$ the remainder is $a$. These cases use the identity directly, without assigning a phase-dual metric or evaluating a zero denominator. If $n=0$ all symbols are functions on a one-point space and the phase is zero for every $t$; the same identity argument applies. For $\kappa=0$ the estimate has no improving frequency order. None of these arguments rescale or replace the original $g$.

For composition we need a parameter version that retains more information before restricting to a diagonal. Let $$c(x,\xi,y,\eta)=a(x,\eta)b(y,\xi),\qquad \mathcal B=\exp(i\langle D_y,D_\eta\rangle)c. \tag{E16}$$ At fixed $(x,\xi)$, let $s=\langle\xi\rangle$ and equip the $(y,\eta)$ variables with $$G_{(y,\eta)}(v,w)=s^{2\delta}|v|^2+ \langle\eta\rangle^{-2\rho}|w|^2. \tag{E17}$$ The evaluation points needed for composition have $\eta=\xi$. At these points the Gauss parameter for the phase $p_y\cdot p_\eta$ is $s^{-\kappa}/2$. We verify the uniform global hypotheses for that evaluation instead of assuming uncertainty at every $\eta$.

Write $r=\langle\eta\rangle$, $d=|\eta-\xi|$. The phase-dual distance based at $(y,\eta)$ includes $4s^{-2\delta}d^2$. Both ratios $r/s$ and $s/r$ are bounded by a fixed power of $1+s^{-2\delta}d^2$, with constants depending only on $\delta<1$. If $r$ and $s$ are within a factor two this is immediate. If $r>2s$, then $d>r/2$ and $s^{-2\delta}d^2\geq (r/s)^2/4$, since $s^{2-2\delta}\geq1$. If $s>2r$, then $d>s/2$, so the same expression is at least $s^{2(1-\delta)}/4$; choosing a sufficiently large power bounds $s/r\leq s$. These inequalities control the ratio of the two frequency coefficients of (E17), every real power weight $r^q$, and their reciprocals. The base coefficient is constant. Slow variation follows from $\rho\leq1$ and the Lipschitz inequality for $\langle\eta\rangle$; the same comparison proves local continuity of $r^q$. Thus every condition of the observation-point form of Sections 7–8 of Quadratic Fourier multipliers at a moving scale holds, with constants independent of $x,\xi,y$.

The tensor $c$ has weight $r^{m_1}s^{m_2}$ in (E17): each $y$-derivative costs at most $s^\delta$, and each $\eta$-derivative costs at most $r^{-\rho}$. Applying the Gauss theorem and then restricting $y=x,\eta=\xi$ gives a remainder of order $m_1+m_2-N\kappa$. To obtain every derivative estimate, differentiate (E16) first. Derivatives commute with its constant-coefficient multiplier. For example $\partial_\xi^\alpha\partial_x^\beta \partial_\eta^{\alpha'}\partial_y^{\beta'}c$ has weight $r^{m_1-\rho|\alpha'|+\delta|\beta|} s^{m_2-\rho|\alpha|+\delta|\beta'|}$. Apply the same theorem with those modified real exponents. Finally, differentiation of the diagonal restriction is the sum of the corresponding derivatives in the two sets of variables. Each term has the required total order in (E2). The product rule bounds every input seminorm by a finite sum of products of seminorms of $a$ and $b$.

For the classical case $(\rho,\delta)=(1,0)$, this argument holds at every $(y,\eta)$, not just $\eta=\xi$, because (E17) no longer depends on $s$. It gives the stronger, pre-diagonal estimate $$\begin{split} &\left\|\partial_\xi^\alpha\partial_x^\beta \partial_\eta^{\alpha'}\partial_y^{\beta'} \left(\mathcal B- \sum_{|\gamma|<N}\frac{\partial_\eta^\gamma a(x,\eta) D_y^\gamma b(y,\xi)}{\gamma!}\right)\right\|\\ &\hspace{12mm}\leq C \langle\eta\rangle^{m_1-N-|\alpha'|} \langle\xi\rangle^{m_2-|\alpha|}. \end{split} \tag{E18}$$ In particular the loss is attached to the transformed frequency variable before evaluation. The constants in (E18) also involve only finitely many source seminorms.

All these Gauss limits agree with the distributional multipliers wherever both are defined. Multiply the symbols by cutoffs in both variables that tend to one. They stay in bounded symbol sets and converge locally smoothly; their polynomial growth gives convergence in $\mathcal S'$. The distributional multiplier is continuous there, while the Gauss estimate gives the same locally smooth limit on its evaluation set. This also justifies differentiating with respect to the fixed parameters in (E16): differentiated approximants have the estimates just proved, converge locally, and their limits are successive derivatives by the fundamental theorem of calculus along parameter segments.

5. Adjoints and composition

Theorem. For $\delta\leq\rho$, the adjoint symbol (E13) belongs to $S^m_{\rho,\delta}$, and left quantization extends continuously to $\mathcal S'$ by transposition of its adjoint’s Schwartz action. If $a\in S^{m_1}_{\rho,\delta}$ and $b\in S^{m_2}_{\rho,\delta}$, then $$\operatorname{Op}(a)\operatorname{Op}(b)=\operatorname{Op}(a\circ b), \quad a\circ b=\mathcal B(x,\xi,x,\xi) \in S^{m_1+m_2}_{\rho,\delta}. \tag{E19}$$ The identity holds on both $\mathcal S$ and $\mathcal S'$. The product is continuous bilinear. For every $N\geq0$, $$a^\dagger-\sum_{|\alpha|<N} \frac{\partial_\xi^\alpha D_x^\alpha(a^*)}{\alpha!} \in S^{m-N\kappa}_{\rho,\delta}, \tag{E20}$$ $$a\circ b-\sum_{|\alpha|<N} \frac{(\partial_\xi^\alpha a)(D_x^\alpha b)}{\alpha!} \in S^{m_1+m_2-N\kappa}_{\rho,\delta}. \tag{E21}$$ Each remainder map has finite-seminorm control as described above. Adjoint transformation sends bounded sequences converging locally smoothly to sequences with the same property; the product has the analogous bounded-set continuity in both factors.

Proof. The symbol assertions and their estimates follow from (E15)–(E18), including $N=0$. Formula (E13) and the Schwartz action prove extension to tempered distributions. For Schwartz symbols and $u\in\mathcal S$, Fubini’s theorem applied to two instances of (E10) gives $$(a\circ b)(x,\xi)=(2\pi)^{-n} \iint e^{i(x-y)\cdot(\eta-\xi)}a(x,\eta)b(y,\xi)\,dy\,d\eta, \tag{E22}$$ which equals (E16) restricted to the diagonal. The sign can also be checked by taking $a=a(\xi)$, $b=e^{iv\cdot x}b_0(\xi)$: (E22) becomes $e^{iv\cdot x}a(\xi+v)b_0(\xi)$, agreeing with (E21).

Choose compactly supported smooth approximants $a_j,b_j$ as at the end of Section 4. Their symbols and products remain bounded in their respective orders and converge locally smoothly. For any fixed $u\in\mathcal S$, (E10) and dominated convergence give local smooth convergence of the outputs. The Schwartz estimates of Section 3 bound all output Schwartz seminorms uniformly. Local convergence plus those stronger uniform seminorms gives convergence in every Schwartz seminorm: outside a large base ball use one extra power of $\langle x\rangle$, and inside use local convergence. Thus $\operatorname{Op}(a_j)u\to\operatorname{Op}(a)u$ in $\mathcal S$, and similarly for $b_j$ and their products. Boundedness of the Schwartz operator seminorms allows passage through the double composition. This proves (E19) on $\mathcal S$.

Taking adjoints there identifies the transpose of the product with the reverse product of the transposes. Apply their continuous extensions to $\mathcal S'$ and test against $\mathcal S$ to obtain (E19) on $\mathcal S'$. Equivalently, the identities of pairings determine the extension uniquely. Kernel injectivity proves associativity of $\circ$ and that the constant symbol $I$ is its unit. ∎

When $\delta<\rho$, (E20)–(E21) are asymptotic expansions in decreasing orders. When $\delta=\rho<1$, closure and every finite remainder estimate survive, but all their target orders are the original order. Taking more terms then supplies no smoothing assertion. This distinction will be used in the inverse construction.

6. A finite-derivative $L^2$ estimate

We first prove an estimate that involves no symbol order. Suppose $a(x,\xi)$ is smooth and all derivatives up to a sufficiently large fixed order are bounded. Then $$\|\operatorname{Op}(a)u\|_2\leq C_n M\|u\|_2, \qquad M=\max_{|\alpha|+|\beta|\leq L_n} \|\partial_\xi^\alpha\partial_x^\beta a\|_\infty. \tag{E23}$$ The number $L_n$ is finite; the proof permits $L_n=4N$ for any integer $N>n/2$. Only this estimate, not an optimal derivative count, is needed.

We record the integral estimate used in its proof. If a measurable kernel $K(s,t)$ on any two copies of a measure space satisfies $\sup_t\int|K(s,t)|\,ds\leq A$ and $\sup_s\int|K(s,t)|\,dt\leq B$, then its canonical pairing operator has $L^2$ norm at most $\sqrt{AB}$. The raw integral agrees globally under semifinite output measure, as proved below. Pointwise Cauchy–Schwarz gives $$\left|\int K(s,t)u(t)\,dt\right|^2 \leq\left(\int|K(s,t)|\,dt\right) \left(\int|K(s,t)||u(t)|^2\,dt\right). \tag{E24}$$ On semifinite output measures, integration in $s$ on finite pieces by Tonelli and the global recovery below give $AB\|u\|_2^2$. The raw integral therefore supplies the bounded extension there. This includes continuous kernels with equal marginal bound $A=B=C$ and norm at most $C$.

The exact measure-space interpretation. For arbitrary measure spaces, the operator in the preceding statement is the canonical pairing operator, not an unrestricted raw-integral operator. Write $\mu$ for the output measure in $s$ and $\nu$ for the input measure in $t$. Their original constants are column $A$ and row $B$; this is $A_{\rm SC}=B$, $B_{\rm SC}=A$ in the complete Schur proof, Section1.1. Joint product measurability is required.

For finite-valued representatives $u\in L^2(\nu)$ and $v\in L^2(\mu)$, their actual supports are sigma-finite, since $$S_u=\bigcup_{k\geq1}\{|u|>1/k\},\qquad \nu\{|u|>1/k\}\leq k^2\|u\|_2^2,$$ and the same bound holds for $v$. On the original restricted product $S_v\times S_u$, Tonelli and Cauchy–Schwarz therefore prove $$\begin{split} b_K(u,v)&=\int_{S_v}\int_{S_u} K(s,t)u(t)\overline{v(s)}\,d\nu(t)\,d\mu(s),\\ \int_{S_v\times S_u}|K|\,|u|^2\,d(\mu\otimes\nu)&\leq A\|u\|_2^2,\\ \int_{S_v\times S_u}|K|\,|v|^2\,d(\mu\otimes\nu)&\leq B\|v\|_2^2,\\ |b_K(u,v)|&\leq\sqrt{AB}\|u\|_2\|v\|_2. \end{split} \tag{E24a}$$ Absolute convergence also proves invariance under changing the original null representatives and sesquilinearity. The complete Hilbert representation argument SC10–SC13 gives the unique operator $T_K$ with $\langle T_Ku,v\rangle=b_K(u,v)$ and $\|T_K\|\leq\sqrt{AB}$, including both zero-constant cases.

SC14–SC23 prove the raw integral’s measurability and existence on every finite output piece, the exact finite-piece comparison with $T_Ku$, and the onto isometry $L^2(\mu)\to F_\mu/N_\mu$. The raw representative has class $i_\mu(T_Ku)$. Under semifinite output measure SC24 proves that the divergent set is globally null and that this raw representative equals $T_Ku$ in the original global $L^2(\mu)$. The explicit SC2–SC6 counterexample shows why truncation alone cannot establish this global assertion on arbitrary output measures.

In every application below the measure is Lebesgue measure, or its original multiple $d\mu(q,p)=(2\pi)^{-n}dq\,dp$, hence is sigma-finite and semifinite. Indeed the boxes $[-k,k]^{2n}$ have their exact finite phase-space measures $(2\pi)^{-n}(2k)^{2n}$ and exhaust the space. A measurable set with positive measure then has a positive finite-measure intersection with one of these boxes. The same proof applies to the ordinary position and frequency measures. Thus the raw integral and all bounds below have the proved meaning. The finite matrix case keeps the original operator norm and both vector dimensions, exactly as SC26 proves.

Proof of (E23). Fix $\phi\in\mathcal S(\mathbb R^n)$ with $\|\phi\|_2=1$, and put $\phi_{q,p}(x)=e^{ip\cdot x}\phi(x-q)$. Use phase-space measure $d\mu(q,p)=(2\pi)^{-n}dq\,dp$. The packet transform $Vu(q,p)=(u,\phi_{q,p})$ is an isometry from $L^2$ into $L^2(d\mu)$: Plancherel in $p$, followed by integration in $q$, gives $$\int|Vu(q,p)|^2d\mu(q,p) =\iint |u(x)|^2|\phi(x-q)|^2\,dx\,dq=\|u\|_2^2. \tag{E25}$$ The synthesis map $V^*f=\int f(Q)\phi_Q\,d\mu(Q)$, initially for bounded compactly supported $f$, has norm at most one: pairing with an $L^2$ function, applying Cauchy–Schwarz and (E25), and taking the supremum over unit vectors proves this bound. Completeness extends synthesis to every $f\in L^2(d\mu)$. Polarization of (E25) gives $V^*V=I$. For Schwartz $u$, $Vu$ decreases rapidly in both packet variables; integration by parts in $x$ controls the momentum variable and the product of two Schwartz functions controls the position variable. Thus the reconstruction $u=\int Vu(Q)\phi_Q\,d\mu(Q)$ converges in $\mathcal S$.

Consider the matrix of $A=\operatorname{Op}(a)$ between two packets, with output packet $(q,p)$ and input packet $(q',p')$. Substitute $x=q+s$, $\xi=p'+t$ in (E10). The integral below is the amplitude of this matrix; its full scalar phase is computed immediately afterward: $$(2\pi)^{-n}\iint e^{i[s\cdot(p'-p)+t\cdot(q-q')]} e^{is\cdot t}a(q+s,p'+t)\widehat\phi(t)\overline{\phi(s)},ds\,dt. \tag{E26}$$

Keep the original $\phi_{q,p}(x)=e^{ip\cdot x}\phi(x-q)$, its $L^2$ norm one, $d\mu(q,p)=(2\pi)^{-n}dq\,dp$, and the linear-first pairing. Directly changing variables in the original forward transform gives

$$\widehat{\phi_{q',p'}}(\xi) =e^{-iq'\cdot(\xi-p')}\widehat\phi(\xi-p'). \tag{EC12}$$

Insert this expression into E10 and pair with $\phi_{q,p}$. With $x=q+s$ and $\xi=p'+t$, the full phase before any grouping is

$$(q+s)\cdot(p'+t)-q'\cdot t-p\cdot(q+s) =q\cdot(p'-p)+s\cdot(p'-p) +t\cdot(q-q')+s\cdot t. \tag{EC13}$$

Consequently the actual original packet matrix is

$$\begin{split} (\operatorname{Op}(a)\phi_{q',p'},\phi_{q,p}) &=e^{iq\cdot(p'-p)}(2\pi)^{-n} \iint e^{i[s\cdot(p'-p)+t\cdot(q-q')]}\\ &\quad{}\times e^{is\cdot t} a(q+s,p'+t)\widehat\phi(t)\overline{\phi(s)}\,ds\,dt . \end{split} \tag{EC14}$$

The integration measure is $ds\,dt$: the comma before $ds$ in the previous E26 is typographical punctuation, not an amplitude factor. All scalar phase factors, the original inverse Fourier multiplier and the matrix order remain. The omitted factor has absolute value one for these real packet parameters, so E27’s absolute-value estimate and both Schur marginals remain valid. It is independent of $s,t$ and therefore survives every displayed integration by parts unchanged.

Apply $(1-\Delta_s)^N(1-\Delta_t)^N$ to the last three factors by integration by parts. Derivatives of $e^{is\cdot t}$ create only polynomials in $s,t$; derivatives of the two fixed Schwartz functions absorb those polynomials. Consequently the absolute integral of all differentiated amplitudes is bounded by $C_{n,N,\phi}M$, uniformly in all four packet parameters. The derivative order of $a$ is at most $4N$. We obtain $$|(A\phi_{q',p'},\phi_{q,p})| \leq CM\langle q-q'\rangle^{-2N} \langle p-p'\rangle^{-2N}. \tag{E27}$$ Since $2N>n$, both phase-space marginals are integrable and uniformly bounded by $CM$. Equation (E24) bounds the corresponding operator $\mathcal M$ on $L^2(d\mu)$. The zeroth bound on $a$ already makes (E10) a continuous map $\mathcal S\to\mathcal S'$, since its outputs are bounded functions. For Schwartz $u,v$, their packet reconstructions and (E27) give $(Au,v)=(\mathcal M Vu,Vv)$. All integrals converge by the decay just proved. Hence $A=V^*\mathcal M V$ on $\mathcal S$ as distributions, which establishes both membership of $Au$ in $L^2$ and the norm bound. No $L^2$ boundedness of $A$ was assumed in forming its packet matrix. This proves (E23), and density gives the extension. ∎

The calculation works for fixed-size matrices by estimating the matrix norm in (E26) and applying the scalar majorant to the norm of the vector input. It therefore has no positivity or scalar-commutativity assumption.

Two consequences also isolate the lower-order mechanisms in the classical calculus. If $a\in S^q_{\rho,\delta}$ with $q<-n$, its kernel is an ordinary continuous function. Every $(x-y)^\alpha K_a(x,y)$ is the inverse-frequency transform multiplied by the exact power of $i$ specified below, applied to $\partial_\xi^\alpha a$. That derivative has integrable absolute value uniformly in $x$. Dominated convergence on compact base sets proves continuity. The needed integration by parts uses a cutoff first and then lets it tend to one.

For the original inverse-frequency kernel E12, differentiating $a$ in $\xi$ and integrating by parts gives

$$K_{\partial_\xi^\alpha a}(x,y) =(-i)^{|\alpha|}(x-y)^\alpha K_a(x,y),\qquad (x-y)^\alpha K_a(x,y) =i^{|\alpha|}K_{\partial_\xi^\alpha a}(x,y). \tag{EC15}$$

These are also exact distributional identities: frequency differentiation is multiplication by $-i(x-y)$ under the inverse transform, proved by testing the defining tempered transforms. When $q<-n$, every displayed frequency derivative is integrable uniformly in $x$. For an ordinary integration proof insert a smooth compact cutoff $\chi(\xi/R)$. Each boundary term contains a positive cutoff derivative, at least one factor $R^{-1}$, and lies where $|\xi|$ is of order $R$. The earlier intermediate estimate claimed an absolute bound by a constant times $R^{n+q-|\alpha|}$, or a smaller power; this exponent is valid when $\rho=1$ but is excessive in the general range $0<\rho<1$. The following correction keeps the entire product rule and proves the actual vanishing bound.

The convergence still follows from $q<-n$, with every original derivative factor retained. Choose the original cutoff to equal one on $|\xi|\leq c_0$ and vanish on $|\xi|\geq c_1$, with $0<c_0<c_1$. Take $R\geq\max(1,c_0^{-1})$. The complete derivative is $$\partial_\xi^\alpha\bigl(a(x,\xi)\chi(\xi/R)\bigr) =\sum_{\beta\leq\alpha}{\alpha\choose\beta} (\partial_\xi^{\alpha-\beta}a)(x,\xi) R^{-|\beta|}(\partial^\beta\chi)(\xi/R). \tag{EC16}$$ For each $0<\beta\leq\alpha$, put $s_\beta=q-\rho|\alpha-\beta|<0$. Its original annular support lies in $c_0R\leq|\xi|\leq c_1R$. On this set $\langle\xi\rangle^{s_\beta}\leq c_0^{s_\beta}R^{s_\beta}$. Equation E2 and the original coordinate volume therefore give, with $\omega_n$ the volume of the Euclidean unit ball, $$\begin{split} &\int\left\|{\alpha\choose\beta} (\partial_\xi^{\alpha-\beta}a)(x,\xi) R^{-|\beta|}(\partial^\beta\chi)(\xi/R)\right\|\,d\xi\\ &\qquad\leq{\alpha\choose\beta}\, p_{q,|\alpha-\beta|}(a)\, \|\partial^\beta\chi\|_\infty\, \omega_nc_1^n c_0^{q-\rho|\alpha-\beta|} R^{n+q-\rho|\alpha-\beta|-|\beta|}. \end{split} \tag{EC17}$$ Every factor is retained, and the bound is uniform in the original base point. Because $\rho>0$ and $|\beta|\geq1$, every exponent is at most $n+q-1<0$. Thus every boundary term tends to zero. The term $\beta=0$ converges in absolute integral by dominated convergence against the integrable original $\partial_\xi^\alpha a$. The integration-by-parts phase has absolute value one, so these are sufficient absolute bounds for that passage. This proves EC15 as an ordinary integral with its exact power of $i$. When $|\alpha|=0$, there are no boundary terms; in dimension zero there is no positive frequency multiindex. The resulting absolute kernel and norm conclusions remain unchanged.

Here is an exact counterexample to the excessive earlier absolute exponent. Take $n=1$, $0<\rho<1$, $\delta=0$, and $q<-1$. Choose smooth $\theta$ equal to zero for $\xi\leq1$ and one for $\xi\geq2$. Define $a(\xi)=0$ for $\xi\leq1$ and $a(\xi)=\theta(\xi)\xi^q e^{i\xi^{1-\rho}}$ for $\xi>1$. It is smooth at the join, and for $\xi\geq2$ every derivative has the full finite form $$a^{(j)}(\xi)=e^{i\xi^{1-\rho}} \sum_{k=0}^j C_{j,k}\xi^{q-j+k(1-\rho)}, \qquad C_{0,0}=1, \tag{EC18}$$ where coefficients outside $0\leq k\leq j$ are zero and direct differentiation gives $$C_{j+1,k}=\bigl(q-j+k(1-\rho)\bigr)C_{j,k} +i(1-\rho)C_{j,k-1}. \tag{EC19}$$ Each exponent is at most $q-\rho j$. On the compact transition interval all terms of the product with $\theta$ are bounded with the original binomial coefficients; there are no position derivatives. Hence $a\in S^q_{\rho,0}$ with the original E2 seminorms. On the positive tail, $$|a'(\xi)|=\left|q\xi^{q-1}+i(1-\rho)\xi^{q-\rho}\right| \geq(1-\rho)\xi^{q-\rho}. \tag{EC20}$$ A smooth compact cutoff which equals one near zero must have a nonzero derivative on some positive interval; continuity supplies a closed interval $J\subset(0,\infty)$ on which $|\chi'|$ is bounded below by a positive constant. For $R$ large enough that $RJ\subset[2,\infty)$, the original $\alpha=2,\beta=1$ boundary term satisfies $$\int_{RJ}\left|2a'(\xi)R^{-1}\chi'(\xi/R)\right|\,d\xi \geq2(1-\rho)R^{q-\rho} \int_J t^{q-\rho}|\chi'(t)|\,dt>0. \tag{EC21}$$ The last integral is a fixed positive number. Division by the earlier proposed power $R^{q-1}$ gives growth by $R^{1-\rho}$, so that stronger absolute bound fails. This is a lower bound for the absolute integral of this particular full product-rule term; it does not assert a lower bound for its oscillatory integral or for a sum with cancellation. The corrected EC17 bound still tends to zero and proves the unchanged original identity and its receiving estimates.