Mixed symbols on every real two-parameter Sobolev scale

This lesson proves continuity on the full two-parameter Sobolev scale. The orders are arbitrary real numbers, the base variable ranges over all of $\mathbb R^n$, and the symbol can be a rectangular matrix between fixed finite-dimensional Hermitian spaces. The original weights $1+|\xi|$ and $1+|\xi'|$ remain explicit.

1. Objects, conventions, and the exact theorem

Fix $n\geq1$ and finite-dimensional complex Hermitian spaces $E,F$. Inner products are linear in their first argument; matrix norms are the induced operator norms. Write $$\xi=(\xi',\xi_n)\in\mathbb R^{n-1}\times\mathbb R, \quad R=1+|\xi|,\quad T=1+|\xi'|, \quad R_0=(1+|\xi|^2)^{1/2},\quad T_0=(1+|\xi'|^2)^{1/2}. \tag{MSB1}$$ When $n=1$, $\xi'$ has no coordinates and $T=T_0=1$. Set $$w_{p,q}(\xi)=R_0(\xi)^pT_0(\xi')^q\qquad(p,q\in\mathbb R). \tag{MSB2}$$ The Fourier convention, including its coefficient, is $$\widehat u(\xi)=\int_{\mathbb R^n}e^{-ix\cdot\xi}u(x)\,dx, \qquad u(x)=(2\pi)^{-n}\int_{\mathbb R^n}e^{ix\cdot\xi}\widehat u(\xi)\,d\xi, \qquad D_j=-i\partial_j. \tag{MSB3}$$ For a smooth $a(x,\xi)\in\operatorname{Hom}(E,F)$, define $$p_{m,m',L}(a)= \max_{|\alpha|+|\beta|\leq L}\sup_{x,\xi} \frac{\|\partial_x^\beta\partial_\xi^\alpha a(x,\xi)\|} {R^{m-\alpha_n}T^{m'-|\alpha'|}}. \tag{MSB4}$$ The class $S^{m,m'}$ consists exactly of those smooth symbols for which every displayed seminorm is finite. There is no restriction of $x$ to a compact set. Matrix factors are multiplied in their given order.

For $p,q\in\mathbb R$, define $H_{(p,q)}(\mathbb R^n;E)$ to be the tempered distributions whose Fourier transform is represented by a locally square-integrable function and for which $$\|u\|_{(p,q)}^2=(2\pi)^{-n} \int_{\mathbb R^n}R_0^{2p}T_0^{2q}\|\widehat u(\xi)\|_E^2\,d\xi<\infty. \tag{MSB5}$$ The proof below shows directly that these are complete Hilbert spaces, that Schwartz functions are dense, and that this definition is equivalent to requiring $w_{p,q}\widehat u\in L^2$ distributionally. The density and distributional conditions therefore do not hide an additional domain assumption.

Theorem. For every $s,t,m,m'\in\mathbb R$ and every $a\in S^{m,m'}(\operatorname{Hom}(E,F))$, the left operator $$A u(x)=\operatorname{Op}(a)u(x) =(2\pi)^{-n}\int e^{ix\cdot\xi}a(x,\xi)\widehat u(\xi)\,d\xi \qquad(u\in\mathcal S(\mathbb R^n;E)) \tag{MSB6}$$ has a unique bounded extension $$A:H_{(s+m,t+m')}(\mathbb R^n;E) \longrightarrow H_{(s,t)}(\mathbb R^n;F). \tag{MSB7}$$ It agrees with the already defined tempered-distribution action of the same left operator. There exist a finite integer $J$ and a constant $C$, depending only on $n,s,t,m,m'$ and the fixed conventions of the course composition and packet estimates, such that $$\|Au\|_{(s,t)}\leq C\,p_{m,m',J}(a)\|u\|_{(s+m,t+m')}. \tag{MSB8}$$ The constant is independent of $a,u$, and of the base point. The operator-norm proofs used below apply to rectangular matrices without a change in this constant when the coefficient-space norms are fixed Hermitian norms; an entrywise proof would instead give a permissible fixed dimension factor. No infinite-dimensional coefficient-space assertion is needed here.

2. Exact bracket correspondence, including negative orders

For $r\geq0$, $1+r^2\leq(1+r)^2\leq2(1+r^2)$. The second difference is $(r-1)^2$. Hence $R/R_0,T/T_0\in[1,\sqrt2]$. Put $z_+=\max(z,0)$, $z_-=\max(-z,0)$. For arbitrary real $u,v$, taking positive powers or reciprocals as appropriate proves $$2^{-(u_-+v_-)/2}R_0^uT_0^v \leq R^uT^v \leq2^{(u_++v_+)/2}R_0^uT_0^v. \tag{MSB9}$$ For a derivative $\alpha$, use exactly $u=m-\alpha_n$, $v=m'-|\alpha'|$. Dividing by these positive weights gives the two continuous identity maps between the original symbol seminorms and the bracket seminorms: $$\begin{split} p^{R,T}_{m,m';\alpha,\beta}(a) &\leq2^{((m-\alpha_n)_-+(m'-|\alpha'|)_-)/2} p^{R_0,T_0}_{m,m';\alpha,\beta}(a),\\ p^{R_0,T_0}_{m,m';\alpha,\beta}(a) &\leq2^{((m-\alpha_n)_++(m'-|\alpha'|)_+)/2} p^{R,T}_{m,m';\alpha,\beta}(a). \end{split} \tag{MSB10}$$ They are identity maps on the actual smooth functions. They preserve the derivatives, base and frequency coordinates, matrix products, and left operator (MSB6); no operation is conjugated or rescaled in this correspondence.

There is an equally exact statement for Sobolev norms. On distributions whose Fourier transform has the function representative in (MSB5), define the additional norm $$\|u\|_{(p,q);R,T}^2=(2\pi)^{-n}\int R^{2p}T^{2q}\|\widehat u\|^2\,d\xi.$$ Multiplication of (MSB9), with $u=p,v=q$, by $\|\widehat u\|$, followed by the squared integral and square root, gives $$2^{-(p_-+q_-)/2}\|u\|_{(p,q)} \leq\|u\|_{(p,q);R,T} \leq2^{(p_++q_+)/2}\|u\|_{(p,q)}. \tag{MSB11}$$ Thus the two norm presentations have exactly the same vectors. The norm with the original weights does not require multiplying an arbitrary distribution by the nonsmooth function $1+|\xi|$: it is defined on the already identified Fourier function representatives. The smooth bracket multipliers used below provide the distributional construction. When $n=1$, all constants from $q$, $m'$, or tangential derivative weights in (MSB9)–(MSB11) may be omitted because the corresponding ratio is one.

3. Smooth multiplier symbols with finite explicit bounds

We first prove a derivative formula valid for every real exponent. In dimension $d\geq1$, put $h_d(\zeta)=(1+|\zeta|^2)^{1/2}$, $z=\zeta/h_d(\zeta)$. For each multiindex $\gamma$ and real $p$, there is a polynomial $P_{\gamma,p}^{(d)}(z)$ such that $$\partial_\zeta^\gamma h_d(\zeta)^p =h_d(\zeta)^{p-|\gamma|}P_{\gamma,p}^{(d)}(\zeta/h_d(\zeta)). \tag{MSB12}$$ Here is a finite, exact construction, which also supplies constants. Start with $P_{0,p}^{(d)}=1$, and choose a fixed coordinate order for the $|\gamma|$ differentiations, with each coordinate repeated its specified number of times. Given the polynomial after $k$ derivatives, differentiation in coordinate $j$ replaces it by $$P\longmapsto(p-k)z_jP +\sum_{\ell=1}^d(\delta_{j\ell}-z_jz_\ell)\partial_{z_\ell}P. \tag{MSB13}$$ Indeed $\partial_jh_d=z_j$, and $\partial_jz_\ell=h_d^{-1}(\delta_{j\ell}-z_jz_\ell)$. The product and chain rules therefore prove (MSB12) at the next step, including its coefficient $p-k$. This inductive construction works for negative, zero, and nonintegral $p$ without alteration. Different orders of differentiation give the same value on the displayed arguments because the actual smooth mixed derivatives commute; a single fixed construction is enough for the bounds.

Let $C^{(d)}_{p,\gamma}$ be the sum of the absolute values of all coefficients of this polynomial. Since every coordinate of $z$ has modulus at most one, $$|\partial^\gamma h_d^p| \leq C^{(d)}_{p,\gamma}h_d^{p-|\gamma|}. \tag{MSB14}$$ For $d=0$, define $h_0=1$, take the sole empty multiindex, and set $C^{(0)}_{p,0}=1$. This states exactly what happens to the tangential factor when $n=1$.

Leibniz differentiation of $w_{p,q}=R_0^pT_0^q$ has only terms where all normal derivatives fall on $R_0^p$. Thus $$\partial_\xi^\alpha w_{p,q} =\sum_{\gamma'\leq\alpha'} \binom{\alpha'}{\gamma'} (\partial_\xi^{(\gamma',\alpha_n)}R_0^p) (\partial_{\xi'}^{\alpha'-\gamma'}T_0^q). \tag{MSB15}$$ Using (MSB14) on both factors, each term is bounded by its coefficient constant times $$R_0^{p-\alpha_n-|\gamma'|} T_0^{q-|\alpha'|+|\gamma'|} =R_0^{p-\alpha_n}T_0^{q-|\alpha'|} (T_0/R_0)^{|\gamma'|} \leq R_0^{p-\alpha_n}T_0^{q-|\alpha'|}.$$ The inequality uses $1\leq T_0\leq R_0$ and does not depend on signs of $p,q$. Set $$K_{p,q,\alpha}= \sum_{\gamma'\leq\alpha'}\binom{\alpha'}{\gamma'} C^{(n)}_{p,(\gamma',\alpha_n)} C^{(n-1)}_{q,\alpha'-\gamma'},$$ and $$K_{p,q,L}^{R,T}= \max_{|\alpha|\leq L} 2^{((p-\alpha_n)_-+(q-|\alpha'|)_-)/2}K_{p,q,\alpha}. \tag{MSB16}$$ Equations (MSB10), (MSB15), and the vanishing base derivatives now prove $$w_{p,q}I_E\in S^{p,q}(\operatorname{End}E), \qquad p_{p,q,L}(w_{p,q}I_E)\leq K_{p,q,L}^{R,T}. \tag{MSB17}$$ The same proof holds with $E$ replaced by $F$, and all real orders, including $(-p,-q)$, are covered. These are explicit finite derivative constants, not merely a formal assertion that order-reducing multipliers belong to a symbol class.

Every derivative of $w_{p,q}$ and its reciprocal grows at most polynomially. For example the bracket estimate just proved is bounded above by $K_{p,q,\alpha}R_0^{|p|+|q|+|\alpha|}$, a deliberately nonsharp bound sufficient here. Leibniz differentiation then proves that multiplication by either function preserves Schwartz space continuously: for a fixed seminorm $\sup_\xi|\xi^\nu\partial^\alpha(w_{p,q}v)|$, finitely many derivatives of $v$, with finitely increased polynomial weights, bound every term. Both multipliers consequently act continuously on tempered distributions by transposition. Their pointwise product is one, so those operations are inverse maps there as well as on Schwartz space.

4. Hilbert spaces, exact multiplier domains, and density

Define the Fourier multiplier $$J_{p,q}u=\mathcal F^{-1}(w_{p,q}\widehat u). \tag{MSB18}$$ The preceding proof makes it a continuous automorphism of both $\mathcal S$ and $\mathcal S'$, with inverse $J_{-p,-q}$. Products satisfy $$J_{p,q}J_{r,v}=J_{p+r,q+v}\quad\hbox{on all of }\mathcal S'. \tag{MSB19}$$ This identity follows from the literal product of the two smooth positive Fourier multipliers, so no unbounded-operator product on an unstated $L^2$ domain is being taken.

If $u\in H_{(p,q)}$, Plancherel in the convention (MSB3) gives $$\|J_{p,q}u\|_{L^2}^2=(2\pi)^{-n}\int|w_{p,q}|^2\|\widehat u\|^2 =\|u\|_{(p,q)}^2. \tag{MSB20}$$ Conversely, given $v\in L^2$, the measurable function $f=w_{-p,-q}\widehat v$ is locally in $L^2$ and defines a tempered distribution. In fact $w_{-p,-q}\leq R_0^{|p|+|q|}$; hence, for any Schwartz function $\psi$, $$\left|\int\langle f(\xi),\psi(\xi)\rangle\,d\xi\right| \leq\|\widehat v\|_2\|R_0^{|p|+|q|}\psi\|_2. \tag{MSB21}$$ The right side is bounded by a Schwartz seminorm: choose any integer $M>|p|+|q|+n/2$, factor out $\sup R_0^M\|\psi\|$, and integrate $R_0^{-2(M-|p|-|q|)}$. That integral is finite, as follows by splitting into the unit ball and dyadic shells with volumes at most a fixed dimensional constant times $2^{kn}$. Taking the inverse Fourier transform gives $u=J_{-p,-q}v\in H_{(p,q)}$. Thus $$J_{p,q}:H_{(p,q)}\longrightarrow L^2 \quad\hbox{is an onto isometry with inverse }J_{-p,-q}. \tag{MSB22}$$ It follows from the completeness of $L^2$ that the normed space in (MSB5) is complete. The weighted Fourier integral with the Hermitian pairing defines its inner product and is positive definite. The same estimate (MSB21), with $v=J_{p,q}u$, proves continuous embedding $H_{(p,q)}\to\mathcal S'$. In particular its vector is uniquely determined by its distribution, not just by an unspecified completion class.

There is no ambiguity in the alternative distributional definition. If a tempered distribution $u$ satisfies $w_{p,q}\widehat u=g\in L^2$, multiplication by the smooth inverse gives $\widehat u=w_{-p,-q}g$; this is the locally square-integrable representative just used. The weighted integral is finite. Conversely the function representative in (MSB5) immediately gives such an $L^2$ product.

For density, use density of Schwartz functions in $L^2$, one of the explicitly stated Fourier-analysis entry results in Section 6 of From symbol estimates to operators on every Sobolev scale. If $v_j\in\mathcal S$ tends to $J_{p,q}u$ in $L^2$, put $u_j=J_{-p,-q}v_j$. The multiplier bounds prove $u_j\in\mathcal S$, and (MSB20) proves $$\|u_j-u\|_{(p,q)}=\|v_j-J_{p,q}u\|_2\longrightarrow0. \tag{MSB23}$$ This proves density for every pair of real orders without assuming a nonnegative exponent. More generally, (MSB19)–(MSB20) give the full-domain isometric isomorphism $$J_{a,b}:H_{(p,q)}\longrightarrow H_{(p-a,q-b)}, \qquad\|J_{a,b}u\|_{(p-a,q-b)}=\|u\|_{(p,q)}. \tag{MSB24}$$ In each case its domain is the entire displayed input space and its inverse is $J_{-a,-b}$.

5. The actual order-zero theorem used here

The finite-derivative estimate in Section 6 of From symbol estimates to operators on every Sobolev scale, (E23), says the following. If a smooth global left symbol $c(x,\xi)$ has bounded derivatives through total order $L_n=4N$, where $N$ is any fixed integer satisfying $N>n/2$, then $$\|\operatorname{Op}(c)v\|_{L^2(F)} \leq C_{n,N,\phi} \max_{|\alpha|+|\beta|\leq4N} \|\partial_x^\beta\partial_\xi^\alpha c\|_\infty \|v\|_{L^2(E)}. \tag{MSB25}$$ Here $\phi$ is the single fixed unit-norm Schwartz packet window chosen in that proof. After it and $N$ have been fixed from $n$, this is a dimensional constant. The packet proof there, including (E24)–(E27), supplies this estimate. Its hypothesis is a bounded finite collection of derivatives, with no frequency-decay exponent or metric hypothesis.

For precision about that proof’s constants and rectangular extension, its packet kernel has norm at most $C'_{n,N,\phi}M\langle q-q'\rangle^{-2N}\langle p-p'\rangle^{-2N}$, where $M$ is the maximum in (MSB25). Applying $(1-\Delta)^N$ in each of the two packet variables differentiates the symbol at most $4N$ times; all other differentiated factors are fixed Schwartz functions times fixed polynomials. Their absolute integrals define the finite constant $C'_{n,N,\phi}$, which includes the Fourier coefficient $(2\pi)^{-n}$. The two kernel marginals, with packet measure $(2\pi)^{-n}dq\,dp$, are bounded by $$(2\pi)^{-n}C'_{n,N,\phi}M \left(\int_{\mathbb R^n}\langle z\rangle^{-2N}\,dz\right)^2. \tag{MSB26}$$ The integral is finite because $2N>n$. The scalar kernel bound applied to vector norms and the isometric packet transform give (MSB25), with this marginal constant. This is also why rectangular matrices are permitted: every differentiated matrix is bounded in operator norm, and the resulting scalar majorant controls the norm of a vector input. No products of incompatible coefficient spaces or commuting matrix assumption enters this argument.

Now $c\in S^{0,0}$ implies $$\|\partial_x^\beta\partial_\xi^\alpha c(x,\xi)\| \leq p_{0,0,4N}(c)R^{-\alpha_n}T^{-|\alpha'|} \leq p_{0,0,4N}(c) \quad(|\alpha|+|\beta|\leq4N), \tag{MSB27}$$ because $R,T\geq1$. Thus the actual hypotheses of (MSB25) hold. This avoids an invalid claim that the mixed symbol belongs to some classical class with uniform positive isotropic frequency decay; such an inclusion need not hold near fixed tangential frequency and arbitrarily large normal frequency.

6. Exact conjugation and the complete continuity proof

We use precisely the mixed composition already proved in Sections 1 and 6 of Composition of mixed symbols with two different remainder estimates. For compatible rectangular symbols $f\in S^{p,q}$, $g\in S^{r,v}$, it constructs the actual left composition symbol $C_1(f,g)\in S^{p+r,q+v}$, proves $$\operatorname{Op}(f)\operatorname{Op}(g) =\operatorname{Op}(C_1(f,g))\quad\hbox{on }\mathcal S\hbox{ and }\mathcal S', \tag{MSB28}$$ and, for each $L$, proves a finite-seminorm bound $$p_{p+r,q+v,L}(C_1(f,g)) \leq C_{L;p,q,r,v,n}\,p_{p,q,J_L}(f)p_{r,v,J_L}(g). \tag{MSB29}$$ The integer $J_L$ and constant depend only on the indicated fixed orders, dimension, and structural constants established there. The proof checks the precise mixed metric, the quadratic-transform parameter, its finite-seminorm continuity, and the common Schwartz and distribution domains. We use its completed mixed assertion, not an unsupported operator-composition statement for arbitrary general metrics. The lower-order remainder part of that theorem is not required here.

Right composition with a scalar Fourier multiplier is even more direct. Put $$q=w_{-s-m,-t-m'},\qquad d(x,\xi)=a(x,\xi)q(\xi). \tag{MSB30}$$ For $v\in\mathcal S(E)$, the Fourier transform of $J_{-s-m,-t-m'}v$ is exactly $q\widehat v$. Substitution into the absolutely convergent integral (MSB6) gives $$A J_{-s-m,-t-m'}v=\operatorname{Op}(d)v. \tag{MSB31}$$ The same identity holds on $\mathcal S'$: both operators have continuous Schwartz adjoints, by Section 6 of Composition of mixed symbols with two different remainder estimates, so the identity just proved on Schwartz inputs gives equality of their Schwartz adjoints when paired against any second Schwartz vector. Transposing this equality gives the distributional identity. Thus no extension of a product beyond its known distributional domains is assumed.

The derivative product rule, (MSB17), and the exact addition of the two weight exponents give $$d\in S^{-s,-t},\qquad p_{-s,-t,L}(d) \leq2^L p_{m,m',L}(a)K^{R,T}_{-s-m,-t-m',L}. \tag{MSB32}$$ To check the coefficient bound, only the frequency derivatives may split between $a$ and $q$; every base derivative acts on $a$. The sum of the frequency binomial coefficients is $2^{|\alpha|}\leq2^L$. In each summand the $R$ exponent is $m-\gamma_n-s-m-(\alpha_n-\gamma_n)=-s-\alpha_n$, and the $T$ exponent is $-t-|\alpha'|$. This proves exactly the stated mixed order, with no loss.

Apply (MSB28)–(MSB29) to $f=w_{s,t}I_F\in S^{s,t}$ and $g=d\in S^{-s,-t}$. Define $$c=C_1(w_{s,t}I_F,d)\in S^{0,0}.$$ On both $\mathcal S(E)$ and $\mathcal S'(E)$ we have the exact identity $$J_{s,t}A J_{-s-m,-t-m'}=\operatorname{Op}(c). \tag{MSB33}$$ Taking $L=4N$ in (MSB29), and enlarging its finite input order to a common $J$ if necessary, (MSB17) and (MSB32) yield $$p_{0,0,4N}(c) \leq C_{4N;s,t,-s,-t,n}\,2^J K^{R,T}_{s,t,J}K^{R,T}_{-s-m,-t-m',J} p_{m,m',J}(a). \tag{MSB34}$$ Every factor is independent of $a,u$ and finite for arbitrary real orders. This formula states the complete constant dependence in (MSB8): multiply its right-hand prefactor by the fixed packet constant $C_{n,N,\phi}$ in (MSB25). It does not assert an unproved universal optimal numerical constant for either prior theorem.

By (MSB25) and (MSB27), the distribution operator $\operatorname{Op}(c)$ restricts to an everywhere-defined bounded map $B:L^2(E)\to L^2(F)$ with norm at most that prefactor. Its agreement with the distribution action follows by approximating an $L^2$ vector by Schwartz vectors: the bounded extension converges in $L^2$, hence in $\mathcal S'$, whereas the previously proved distributional continuity gives the same distribution limit.

Using the isomorphisms in (MSB22), define on the entire input space in (MSB7) $$A_{s,t}=J_{-s,-t}\, B\, J_{s+m,t+m'}. \tag{MSB35}$$ Each arrow now has a stated full domain and codomain: $$H_{(s+m,t+m')}(E)\xrightarrow{\ J_{s+m,t+m'}\ }L^2(E) \xrightarrow{\ B\ }L^2(F) \xrightarrow{\ J_{-s,-t}\ }H_{(s,t)}(F).$$ The first and last are isometries. Hence (MSB34) and (MSB25) prove (MSB8) for all input vectors. On Schwartz vectors, (MSB33) and the inverse identity (MSB19) show that (MSB35) is exactly $A$.

For an arbitrary $u\in H_{(s+m,t+m')}$, choose the dense Schwartz approximation (MSB23). Then $Au_j\to A_{s,t}u$ in the target space and hence in $\mathcal S'$. Also $u_j\to u$ in $\mathcal S'$ and the distributional action of $A$ is continuous, so $Au_j\to Au$ there. Uniqueness of distributional limits proves $A_{s,t}u=Au$. This both identifies the extension and proves that choices of $s,t$ give the same operator on intersections of their domains. If two bounded extensions existed, their difference would vanish on the dense Schwartz subspace and hence everywhere. Uniqueness follows.

In particular (MSB33) has the full Hilbert-space form $$J_{s,t}A J_{-s-m,-t-m'}:L^2(E)\longrightarrow L^2(F),$$ with domain all of $L^2(E)$, because the middle $A$ acts between exactly the two spaces in (MSB7). One must not instead interpret the display as an unspecified product of three possibly unbounded operators on a single $L^2$ space. The proof has identified every domain and image needed for its meaning.

Finally the statement can be written entirely with the additional original-weight Sobolev norms. Combining (MSB8) and (MSB11) gives $$\|Au\|_{(s,t);R,T} \leq 2^{[s_++t_++(s+m)_-+(t+m')_-]/2} C\,p_{m,m',J}(a)\|u\|_{(s+m,t+m');R,T}. \tag{MSB36}$$ This proves the theorem in both weight presentations with explicit conversion constants and unchanged objects, not just for integer or nonnegative orders.

7. A rectangular multiplier and sharp independent order costs

Let $B:E\to F$ be a fixed nonzero rectangular matrix, and take $$a(x,\xi)=w_{m,m'}(\xi)B. \tag{MSB37}$$ The multiplier proof shows that it belongs to $S^{m,m'}$, with seminorm at most $\|B\|K^{R,T}_{m,m',L}$. The Fourier action and weighted norm are exact: $$\|Au\|_{(s,t)}^2 =(2\pi)^{-n}\int w_{s+m,t+m'}^2\|B\widehat u\|_F^2\,d\xi \leq\|B\|^2\|u\|_{(s+m,t+m')}^2. \tag{MSB38}$$ The operator norm is exactly $\|B\|$. Indeed, in finite dimension the continuous function $v\mapsto\|Bv\|$ attains its maximum on the unit sphere. Choose a unit maximizing vector $v$ and any scalar $h\in C_c^\infty(\mathbb R^n)$ with $(2\pi)^{-n}\int|h|^2=1$. Set $\widehat u=w_{-s-m,-t-m'}h v$. This is a smooth compactly supported Fourier function, hence gives $u\in\mathcal S$. Its input norm is one and (MSB38) gives output norm $\|B\|$.

The two target orders cannot in general be independently increased. For any $\varepsilon>0$, choose the preceding $h$ supported in the ball of radius $1/8$, and set $$h_L(\xi)=h(\xi-Le_n),\qquad \widehat u_L=w_{-s-m,-t-m'}h_Lv,\qquad L\geq1.$$ The original input norm is still one. In the stronger target $H_{(s+\varepsilon,t)}$, the exact cancellation of weights in (MSB38) gives $$\|Au_L\|_{(s+\varepsilon,t)}^2 =(2\pi)^{-n}\|B\|^2\int R_0^{2\varepsilon}|h_L|^2\,d\xi \geq\|B\|^2(L-1/8)^{2\varepsilon}\longrightarrow\infty. \tag{MSB39}$$ Thus an arbitrary positive improvement of the total-frequency target order fails. For $n\geq2$, shift $h$ by $Le_1$ instead and use the stronger target $H_{(s,t+\varepsilon)}$. The same calculation has $T_0^{2\varepsilon}$ in place of $R_0^{2\varepsilon}$, bounded below by $(L-1/8)^{2\varepsilon}$. Hence the tangential order cost is also independently sharp. For $n=1$, the tangential norm factor is exactly one, so no tangential sharpness claim is made and the theorem is independent of $t,m'$.

8. Solved exercise: an exact nonconstant conjugated symbol

Exercise. Fix $v\in\mathbb R^n$, a nonzero rectangular matrix $B:E\to F$, and arbitrary real $m,m',s,t$. For $$a_v(x,\xi)=e^{iv\cdot x}w_{m,m'}(\xi)B,$$ verify its mixed symbol class, compute its exact conjugated order-zero symbol, and compute the operator norm between the spaces in (MSB7). Give a finite explicit upper bound in terms of $v,s,t,B$.

Solution. Differentiation in $x$ multiplies by $(iv)^\beta$, and frequency derivatives act on the multiplier already bounded in Section 3. Thus $$p_{m,m',L}(a_v) \leq\|B\|\max(1,|v|^L)K^{R,T}_{m,m',L}, \tag{MSB40}$$ where the bound $|v^\beta|\leq\max(1,|v|^L)$ holds for $|\beta|\leq L$. Every symbol hypothesis holds globally.

The Fourier transform of multiplication by $e^{iv\cdot x}$ shifts frequency by $v$. Therefore $$\widehat {A_vu}(\eta+v)=w_{m,m'}(\eta)B\widehat u(\eta).$$ For the conjugated operator in (MSB33), this gives the exact left symbol $$c_v(x,\xi)=e^{iv\cdot x} \frac{w_{s,t}(\xi+v)}{w_{s,t}(\xi)}B. \tag{MSB41}$$ One may also recover it by testing (MSB6): the conjugated Fourier action first multiplies the input at $\xi$ by the ratio and then shifts it by $v$. This checks both the sign and the positions of the numerator and denominator. It is not in general equal to the pointwise product $e^{iv\cdot x}B$.

The elementary triangle inequality in $\mathbb R^{n+1}$ gives $R_0(\eta+v)\leq R_0(\eta)+|v|\leq(1+|v|)R_0(\eta)$. Interchanging $\eta$ and $\eta+v$ gives the reciprocal bound. The same reasoning in the tangential variables gives $$(1+|v|)^{-1}\leq\frac{R_0(\eta+v)}{R_0(\eta)}\leq1+|v|, \quad (1+|v'|)^{-1}\leq\frac{T_0(\eta'+v')}{T_0(\eta')}\leq1+|v'|.$$ Taking the actual real powers yields $$r_v(\eta):=\frac{w_{s,t}(\eta+v)}{w_{s,t}(\eta)} \leq(1+|v|)^{|s|}(1+|v'|)^{|t|}. \tag{MSB42}$$ The ratio is positive and continuous. After changing variables by $\eta+v$ in the target norm, one obtains $$\|A_vu\|_{(s,t)}^2 =(2\pi)^{-n}\int r_v(\eta)^2 \|B[w_{s+m,t+m'}(\eta)\widehat u(\eta)]\|^2\,d\eta.$$ Consequently the exact operator norm is $$\|A_v\|_{H_{(s+m,t+m')}\to H_{(s,t)}} =\|B\|\sup_{\eta\in\mathbb R^n}r_v(\eta) \leq\|B\|(1+|v|)^{|s|}(1+|v'|)^{|t|}. \tag{MSB43}$$ The upper bound follows from the integral. To prove equality, choose a maximizing unit vector for $B$, and a point where the ratio is within any prescribed positive error of its supremum. Continuity supplies a ball on which the ratio remains within twice that error. A normalized smooth Fourier bump supported in that ball, multiplied by $w_{-s-m,-t-m'}$ and the chosen vector as in Section 7, has unit input norm and gives the matching lower bound. Let the error decrease to zero. This argument also covers a supremum approached only at infinity. The formula keeps all real orders; the absence of $m,m'$ in the final norm is the proved exact cancellation, not an assumption about their signs.

9. Boundary traces with the two weights retained

Later collar arguments require a trace theorem in these spaces, in addition to operator continuity. Let $k\geq0$ be an integer, $s>k+1/2$, and $t\in\mathbb R$. For a Schwartz vector define $\gamma_{k,a}u=(D_n^ku)|_{x_n=a}$, where $a\in\mathbb R$. Then

$$\gamma_{k,a}:H_{(s,t)}(\mathbb R^n;E) \longrightarrow H^{s+t-k-1/2}(\mathbb R^{n-1};E), \qquad \|\gamma_{k,a}u\|^2 \leq \frac{c_{s,k}}{2\pi}\|u\|_{(s,t)}^2, \tag{MSB44}$$

uniformly in $a$, with the finite, positive constant

$$c_{s,k}=\int_{\mathbb R}\lambda^{2k}(1+\lambda^2)^{-s}\,d\lambda. \tag{MSB45}$$

The target norm uses the Fourier coefficient $(2\pi)^{-(n-1)}$. For $n=1$, the target is $E$, with its given norm. The resulting $E$-valued boundary section depends continuously on $a$ in the displayed target space.

Proof. Write $\xi=(\eta,\zeta)$ and $h=(1+|\eta|^2)^{1/2}=T_0(\eta)$. Fourier inversion in the normal coordinate gives exactly

$$\widehat{\gamma_{k,a}u}(\eta) =(2\pi)^{-1}\int_{\mathbb R} e^{ia\zeta}\zeta^k\widehat u(\eta,\zeta)\,d\zeta. \tag{MSB46}$$

The power $\zeta^k$ has this sign because $D_n=-i\partial_n$. Cauchy–Schwarz with the original factors $(h^2+\zeta^2)^{s/2}h^t$ yields

$$\begin{aligned} \|\widehat{\gamma_{k,a}u}(\eta)\|^2 &\leq (2\pi)^{-2} \left(\int_{\mathbb R} \zeta^{2k}(h^2+\zeta^2)^{-s}h^{-2t}\,d\zeta\right) \left(\int_{\mathbb R} (h^2+\zeta^2)^sh^{2t} \|\widehat u(\eta,\zeta)\|^2\,d\zeta\right),\\ \int_{\mathbb R}\zeta^{2k}(h^2+\zeta^2)^{-s}h^{-2t}\,d\zeta &=h^{2k+1-2s-2t} \int_{\mathbb R}\lambda^{2k}(1+\lambda^2)^{-s}\,d\lambda, \qquad \zeta=h\lambda,\quad d\zeta=h\,d\lambda. \end{aligned} \tag{MSB47}$$

The last integral is finite near zero since $k\geq0$, and on the two real tails since $2k-2s<-1$. Multiply the first bound by the exact target weight $h^{2(s+t-k-1/2)}$ and integrate with coefficient $(2\pi)^{-(n-1)}$. The factor in the second line cancels that target weight, and the remaining coefficient is $(2\pi)^{-(n+1)}c_{s,k}$. Comparing it with the input coefficient $(2\pi)^{-n}$ proves (MSB44), retaining the constant $c_{s,k}/(2\pi)$. When $n=1$, $h=1$ and the tangential integral is over the singleton $\mathbb R^0$ with mass one; the same calculation proves the stated case.

Schwartz density from (MSB23) extends every $\gamma_{k,a}$ uniquely. For a general input its Fourier representative satisfies (MSB46) for almost every $\eta$: (MSB47) proves absolute integrability of the normal integral by Cauchy–Schwarz. If $a\to b$, apply that same bound with $|e^{ia\zeta}-e^{ib\zeta}|^2$ in the first integral. For each $\eta$, dominated convergence makes that integral tend to zero; it is bounded by four times the first integral in (MSB47). After multiplication by the target weight, its product with the second integral is dominated by four times $c_{s,k}$ times the integrable input Fourier density. A second dominated-convergence application proves norm continuity in $a$.

To pass to the half-space, define the restriction space with its actual quotient norm,

$$\begin{aligned} \bar H_{(s,t)}(\mathbb R^n_+;E) &=\{r^+U:U\in H_{(s,t)}(\mathbb R^n;E)\},\\ \|u\|_{\bar H_{(s,t)}}&= \inf_{r^+U=u}\|U\|_{(s,t)} . \end{aligned} \tag{MSB48}$$

The kernel of $r^+$ is closed in the whole-space Hilbert space. Indeed, convergence in that Hilbert norm implies distributional convergence by (MSB21), so vanishing on every test function supported in the open half-space persists under the limit. The closed-subspace quotient theorem in Banach estimates, quotient spaces and compact parameter arguments therefore makes (MSB48) a complete normed restriction space.

If $r^+W=0$, then $\gamma_{k,a}W=0$ for $a>0$. Here is the distributional justification. Pair (MSB46) with an arbitrary Schwartz tangential test vector. Its continuous function of $a$, integrated against a compactly supported smooth normal test function, is exactly the pairing of $D_n^kW$ with the product test function, by the Fourier formula and the weighted Cauchy–Schwarz bound. If the normal test function is supported in $a>0$, this is zero because $W$ vanishes on that open half-space. A continuous scalar function giving zero against all such tests is zero there; otherwise one small neighborhood of a nonzero value, after multiplication by its conjugate phase, would give a nonzero integral against a nonnegative test function. All tangential pairings therefore vanish. Norm continuity as $a\downarrow0$ gives $\gamma_{k,0}W=0$.

Consequently $\gamma_ku=\gamma_{k,0}U$ is independent of the chosen whole-space extension $U$, and taking the infimum in (MSB44) gives

$$\gamma_k:\bar H_{(s,t)}(\mathbb R^n_+;E) \longrightarrow H^{s+t-k-1/2}(\mathbb R^{n-1};E), \qquad \|\gamma_ku\|^2\leq\frac{c_{s,k}}{2\pi} \|u\|_{\bar H_{(s,t)}}^2 . \tag{MSB49}$$

For smooth inputs this is the original normal derivative at the boundary. Restrictions of the dense whole-space Schwartz subspace are dense in (MSB48), since each extension can be approximated before restriction. Thus the theorem defines the same trace by density, without taking the boundary value of an arbitrary distribution. The strict condition $s>k+1/2$, both real weights, the normal derivative convention, and the full restriction domain remain part of the assertion.

9.1. Extending all Cauchy jets with the smallest norm

The separate estimates (MSB44) do not yet answer a useful boundary question: can we prescribe every derivative up to a fixed order, and what is the least possible cost of doing so? The answer retains the interactions between different derivatives. These interactions are the entries of a finite Gram matrix, not independent scalar trace constants.

For this subsection take $E\ne\{0\}$. For the zero space the extension and minimum formulas give the unique zero maps; there is no nonzero-data sharpness assertion. Fix an integer $K\geq0$, $s>K+1/2$, and $t\in\mathbb R$. Put $d=n-1$, $h=(1+|\eta|^2)^{1/2}$, and define

$$\begin{aligned} \mathcal T_K^{s+t} &=\bigoplus_{k=0}^K H^{s+t-k-1/2}(\mathbb R^d;E),\\ \Gamma_Ku&=(\gamma_{0,0}u,\ldots,\gamma_{K,0}u),\\ (G_{s,K})_{bc} &=\int_{\mathbb R}z^{b+c}(1+z^2)^{-s}\,dz \quad(0\leq b,c\leq K). \end{aligned} \tag{JT1}$$

The direct-sum norm is the sum of the squares of the displayed Sobolev norms. The coefficient of each tangential Fourier integral is $(2\pi)^{-d}$. When $d=0$, that integral is over the singleton with mass one, $h=1$, and the target is $E^{K+1}$.

Every integral in (JT1) is absolutely convergent: its highest possible tail exponent is $2K-2s<-1$. Odd entries vanish by the reflection $z\mapsto-z$. For $b+c=2l$, substitution $\tau=z^2$ on both halves of the line gives the exact value

$$\begin{aligned} (G_{s,K})_{bc} &=\int_0^\infty\tau^{l-1/2}(1+\tau)^{-s}\,d\tau =\mathrm B(l+1/2,s-l-1/2),\\ \mathrm B(u,v)&=\int_0^\infty \tau^{u-1}(1+\tau)^{-(u+v)}\,d\tau \quad(u,v>0),\\ a^*G_{s,K}a &=\int_{\mathbb R}(1+z^2)^{-s} \left|\sum_{b=0}^K a_bz^b\right|^2\,dz>0 \quad(a\ne0). \end{aligned} \tag{JT2}$$

The last inequality follows because a nonzero polynomial cannot vanish on every real interval: its highest nonzero coefficient would otherwise vanish after the corresponding number of differentiations. At a point where it is nonzero, continuity supplies an interval of positive integral. Thus the real symmetric matrix $G_{s,K}$ is positive definite and invertible. Write $g_-$ and $g_+$ for its smallest and largest eigenvalues, both positive. The finite-dimensional spectral theorem in Fourier transforms, finite spectra and convex separation applies to this actual matrix. For vector coefficients the same calculation uses $G_{s,K}\otimes I_E$, preserving the given Hermitian norm on $E$.

An explicit right inverse

For $f=(f_0,\ldots,f_K)\in\mathcal T_K^{s+t}$, set

$$\begin{aligned} w_c(\eta)&=h^{-c}\widehat f_c(\eta),\\ \rho_b(\eta)&=\sum_{c=0}^K(G_{s,K}^{-1})_{bc} h^{-c}\widehat f_c(\eta),\\ \widehat{\mathcal E_{s,K}f}(\eta,\zeta) &=2\pi h^{2s-1}(h^2+\zeta^2)^{-s} \sum_{b=0}^K\left(\frac\zeta h\right)^b \sum_{c=0}^K(G_{s,K}^{-1})_{bc} h^{-c}\widehat f_c(\eta). \end{aligned} \tag{JT3}$$

This is the full Fourier formula, with the same normal frequency, derivative sign and quadratic weights as (MSB46). The normal integral of its $k$-th moment is

$$\begin{aligned} &(2\pi)^{-1}\int_{\mathbb R}\zeta^k \widehat{\mathcal E_{s,K}f}(\eta,\zeta)\,d\zeta\\ &\quad=h^{2s-1}\sum_{b=0}^K\rho_b(\eta) \int_{\mathbb R}\zeta^k \left(\frac\zeta h\right)^b (h^2+\zeta^2)^{-s}\,d\zeta\\ &\quad=h^{2s-1}h^{k+1-2s} \sum_{b=0}^K(G_{s,K})_{kb}\rho_b(\eta) =\widehat f_k(\eta), \qquad \zeta=hz,\quad d\zeta=h\,dz. \end{aligned} \tag{JT4}$$

Each moment is absolutely convergent since $k+b\leq2K$. In particular, $\Gamma_K\mathcal E_{s,K}=I$; no compatibility relation between the prescribed derivatives is required.

To prove that (JT3) defines a member of the original volume space, compute its complete norm rather than estimating each summand separately:

$$\begin{aligned} \|\mathcal E_{s,K}f\|_{(s,t)}^2 &=(2\pi)^{-(d+1)}(2\pi)^2 \int_{\mathbb R^d}h^{4s-2+2t} \int_{\mathbb R}(h^2+\zeta^2)^{-s} \left\|\sum_{b=0}^K\left(\frac\zeta h\right)^b \rho_b(\eta)\right\|^2\,d\zeta\,d\eta\\ &=2\pi(2\pi)^{-d}\int_{\mathbb R^d} h^{2s+2t-1}\sum_{b,c=0}^K (G_{s,K}^{-1})_{bc} \langle w_b(\eta),w_c(\eta)\rangle_E\,d\eta,\\ \frac{2\pi}{g_+}\sum_{k=0}^K \|f_k\|_{H^{s+t-k-1/2}}^2 &\leq\|\mathcal E_{s,K}f\|_{(s,t)}^2 \leq\frac{2\pi}{g_-}\sum_{k=0}^K \|f_k\|_{H^{s+t-k-1/2}}^2. \end{aligned} \tag{JT5}$$

Here the Hermitian product retains the lesson’s convention: it is linear in its first entry. Since the Gram matrix and its inverse are real symmetric, the displayed sum is the same real quadratic form in either coordinate order. The inner normal integral before the substitution is nonnegative; Tonelli therefore justifies integrating it first. Substitution $\zeta=hz$ contributes exactly $h^{1-2s}$ and the full matrix $G_{s,K}$. Since $\rho=(G_{s,K}^{-1}\otimes I_E)w$, its quadratic form is the displayed form with $G_{s,K}^{-1}$. The tangential weight for $f_k$ is exactly $h^{2s+2t-1}h^{-2k}$, proving both bounds. The weighted Fourier representative has finite norm; (MSB21) makes it a tempered distribution and (MSB22) places it in $H_{(s,t)}$. Thus $\mathcal E_{s,K}$ is a bounded linear right inverse. Its formula is independent of $t$, as the computation proves, while its domain and both norms still contain the original $t$.