AN-04 proof edition · CC0; exact source credit and separately licensed prerequisites

Boundary operator bounds, conormal action and the residual obstruction

These connected components retain AN03-U032, Totally characteristic operators on the half space, Sections 2–5, Section 6 through Example 6.7, and Sections 7–13. Original author: Claude Opus 5.5 (Anthropic), September 2026; editorial additions: Codex, September 2026. Both were dedicated to the public domain (CC0). Current prerequisite connections and proof clarifications: AN-04 course-writing task and OpenAI Codex, 5 October 2026, also CC0. The selected components retain every mathematical display, the full scalar and finite-matrix hypotheses, and all five original solved exercises.

The approved mathematical antecedent is Hörmander III, 2007 eBook, ISBN 978-3-540-49938-1, Section 18.3. Its use and ordinary citation are valid. Complete proofs are supplied in the components and the exact earlier programme proofs. The earlier linked components retain their individual licences.

The four components, in proof order, are Boundary tests, lacunary symbols and all normal jets, Resolved corner kernels and their exact inverse, Boundary adjoints, complete composition and distributional action, Boundary operator bounds, conormal action and the residual obstruction. Original section and equation numbers are retained across them. Sections 6.8 (polyhomogeneous corner characterization) and 14 (arbitrary positive-order Sobolev loss) are separate unadopted obligations; the theorems below do not substitute for those results or for the global compressed wave-front calculus.

Use the test and symbol component, the corner kernel bound, and the full adjoint, composition and distributional action. The half-space Hilbert companion also supplies the exact closed-subspace projection and full Hilbert antidual representation used in Proposition 10.2. Its one-sided multiplier is a later prerequisite, not a replacement for the order-zero argument below.

10. Boundedness on $L^2$, Sobolev and Besov spaces

Operators of order 0 are bounded on $L^2(\mathbb R^n_+)$. We prove this first. Then we extend it to Sobolev spaces of integer order, using the commutator identities and duality, and to all real orders and all Besov exponents by interpolation.

Boundedness on $L^2$

Theorem 10.1 (Boundedness on $L^2$). If $a\in S^0_{\mathrm{la}}$, then $T_a$ extends to a bounded operator on $L^2(\mathbb R^n_+)$, with norm bounded in terms of finitely many seminorms of $a$.

Proof. Step A (order $-n-2$). For $a\in S^{-n-2}_{\mathrm{la}}$, Proposition 6.4 and the full Lebesgue Schur proof B3 linked in Section 1, give $\|T_au\|_{L^2(\mathbb R^n_+)}\leq C\|u\|_{L^2(\mathbb R^n_+)}$ for $u\in\overline{\mathcal S}(\mathbb R^n_+)$, since $T_au(x)=\int_{\mathbb R^n_+}K_a(x,y)u(y)dy$. Restrictions of Schwartz functions are dense in $L^2(\mathbb R^n_+)$.

Step B (doubling). Suppose every operator with symbol in $S^{-2k}_{\mathrm{la}}$ is bounded, and let $a\in S^{-k}_{\mathrm{la}}$. For $u\in\overline{\mathcal S}(\mathbb R^n_+)$, Theorems 7.3 and 8.1 give $\|T_au\|^2=(T_{a^\dagger}T_au,u)=(T_cu,u)$ with $c=a^\dagger\#a\in S^{-2k}_{\mathrm{la}}$ (writing $\#$ for the composition symbol of Theorem 8.1). So $\|T_au\|^2\leq\|T_c\|\,\|u\|^2$.

Step C. By Steps A–B, operators with symbols in $S^{-k}_{\mathrm{la}}$ are bounded for $k\geq(n+2)/2$, then for $k\geq(n+2)/4$, and so on; after finitely many steps, for every $k>0$.

Step D (order 0). Let $a\in S^0_{\mathrm{la}}$ and $M>\sup|a|$. The function $c_0=(M^2-|a|^2)^{1/2}-M$ lies in $S^0_+$: it is $G(a,\overline a)$ with $G$ smooth on a neighbourhood of the closed range and $G(0)=0$, so by the chain rule every derivative is a sum of products containing at least one derivative of $a$ (or $a$ itself, since $|G(a)|\leq C|a|$), which gives the decay in $x_n$. Let $c=(c_0)_\rho\in S^0_{\mathrm{la}}$ (Lemma 4.4). For $u\in\overline{\mathcal S}(\mathbb R^n_+)$,

using $2\operatorname{Re}(T_cu,u)=(T_{c+c^\dagger}u,u)$, $\|T_cu\|^2=(T_{c^\dagger\#c}u,u)$ and $\|T_au\|^2=(T_{a^\dagger\#a}u,u)$. The symbol $r$ is lacunary. Its leading part, by Theorem 7.3(a) and (8.3), is $2Mc_0+c_0^2+|a|^2$ modulo $S^{-1}_+$ (recall $c-c_0\in S^{-\infty}_+$ and $c_0$ is real). This equals $(M+c_0)^2-M^2+|a|^2=0$. So $r\in S^{-1}_{\mathrm{la}}$, $T_r$ is bounded by Step C, and $\|T_au\|^2\leq(M^2+\|T_r\|)\|u\|^2$. $\square$

Matrix-valued symbols. For $a$ with values in $L(\mathbb C^p,\mathbb C^q)$ take $M>\sup\|a\|$ and

The full square-root series, including its constant term, is

The square root is positive; $c_0$ is its displayed difference from $M I_p$. For $\|A\|\le r<M$, the ordered power series and all its real and imaginary entry derivatives converge uniformly. Its first term is $-A^*A/(2M)$; hence $C_M(0)=0$ and its first derivative at zero is zero. The square root itself equals $M I_p$ there. The product and chain rules, with the uniform derivative bounds on this ball, prove $c_0\in S^0_+$. The leading symbol of $r$, with all identities explicit, is

Thus the same ordered proof applies, with the input and output vector dimensions retained.

For completeness the series assertion just used follows directly from $c_k=\binom{1/2}{k}(-1)^k$. Its recurrence is $(k+1)c_{k+1}=(k-\tfrac12)c_k$, so $|c_k|\le1$ and the series $y(z)=\sum_{k\ge0}c_kz^k$ converges absolutely for $|z|<1$. It satisfies $2(1-z)y'=-y$, by comparing coefficients, and $y(0)=1$. Therefore the power series $h=y^2$ satisfies $(1-z)h'=-h$, $h(0)=1$. Coefficient comparison gives $h=1-z$. For real $0\le z<1$, continuity and $y^2=1-z>0$ give $y(z)>0$. For a matrix $B=A^*A/M^2$ with $\|B\|\le q<1$, absolute operator-norm convergence permits multiplication of the two series and gives $y(B)^2=I-B$; real coefficients make $y(B)$ self-adjoint. The finite-dimensional spectral proof in the earlier stationary-phase foundations identifies its eigenvalues as the positive numbers $y(\lambda)$. A derivative of order $d$ in the real and imaginary entries of $A$ differentiates at most $d$ factors in an ordered product $B^k$; on a fixed smaller ball it is bounded by a constant times $k^d q^{k-d}$ for $k\ge d$, with finitely many initial terms handled separately. The geometric series with this polynomial factor converges. This proves every stated uniform derivative bound and justifies the symbol chain rule for the finite-matrix square root.

The operator norm bound is linear in a finite collection of symbol seminorms, despite the square-root construction. The finite doubling argument needs only finitely many continuous composition and adjoint seminorms. Choose a finite seminorm $p(a)$ dominating all of them and the supremum norm. If $p(a)>0$, apply the construction to $a/p(a)$ with $M=2$; every bound just proved is then uniform, so $\|T_a\|\le C p(a)$. If $p(a)=0$, the supremum bound makes $a=0$. This proves the required homogeneous finite-seminorm continuity, including the rectangular matrix case.

Sobolev spaces on the half space

Let $\dot H_{(s)}(\overline{\mathbb R}{}^n_+)=\{u\in H_{(s)}:\operatorname{supp}u\subset\overline{\mathbb R}{}^n_+\}$, with the norm of $H_{(s)}$, and let $\overline H_{(s)}(\mathbb R^n_+)$ be the space of restrictions, with $\|u\|_{\overline H_{(s)}}=\inf\{\|U\|_{(s)}:U=u\text{ in }\mathbb R^n_+\}$. Define $\dot B^s_{2,p}(\overline{\mathbb R}{}^n_+)$ in the same way as $\dot H_{(s)}$. We prove the three facts about these spaces that we need.

Proposition 10.2 (Sobolev spaces on the half space).

(a) $C_0^\infty(\mathbb R^n_+)$ is dense in $\dot H_{(s)}(\overline{\mathbb R}{}^n_+)$, and $\overline{\mathcal S}(\mathbb R^n_+)$ is dense in $\overline H_{(s)}(\mathbb R^n_+)$, for every real $s$.

(b) The sesquilinear form $(u,v)=(2\pi)^{-n}\int\widehat u\,\overline{\widehat V}\,d\xi$, for $u\in\dot H_{(s)}(\overline{\mathbb R}{}^n_+)$ and $V\in H_{(-s)}$ any extension of $v\in\overline H_{(-s)}(\mathbb R^n_+)$, is well defined. It identifies each of $\dot H_{(s)}(\overline{\mathbb R}{}^n_+)$ and $\overline H_{(-s)}(\mathbb R^n_+)$ isometrically with the antidual of the other. For $u\in\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)$, $v\in\overline{\mathcal S}(\mathbb R^n_+)$ it equals $\int_{\mathbb R^n_+}u\overline v$.

(c) Let $k\geq0$ be an integer. For $u\in\dot H_{(k)}(\overline{\mathbb R}{}^n_+)$, $\|u\|_{(k)}^2=\sum_{|\alpha|\leq k}\frac{k!}{\alpha!(k-|\alpha|)!}\|D^\alpha u\|^2_{L^2}$. For $u\in\overline H_{(k)}(\mathbb R^n_+)$, with $N_k(u)^2=\sum_{|\alpha|\leq k}\|D^\alpha u\|^2_{L^2(\mathbb R^n_+)}$,

Proof. (a) Let $u\in\dot H_{(s)}$. The translates $u_h=u(\cdot-he_n)$, supported in $x_n\geq h$, converge to $u$ in $H_{(s)}$ as $h\downarrow0$, by dominated convergence on the Fourier side. Mollifying with a kernel supported in $\{|x|<h/2\}$ gives smooth functions supported in $x_n\geq h/2$, converging in $H_{(s)}$; these lie in $H_{(\sigma)}$ for every $\sigma$. Cutting off with $\theta(x/R)$ converges in $H_{(k)}$ for integers $k\geq s$ (Leibniz' rule and dominated convergence), hence in $H_{(s)}$. The second statement holds because $\mathcal S$ is dense in $H_{(s)}$ and restriction is continuous and onto.

(b) $H_{(s)}$ and $H_{(-s)}$ are each other's antiduals, isometrically, under this form: Cauchy–Schwarz with the weights $\langle\xi\rangle^{\pm s}$, with equality for $\widehat V=\langle\xi\rangle^{2s}\widehat u$. If $V=0$ in $\mathbb R^n_+$, then $(\varphi,V)=0$ for $\varphi\in C_0^\infty(\mathbb R^n_+)$, and by (a) $(u,V)=0$ for all $u\in\dot H_{(s)}$; so the form is well defined, and the annihilator of $\dot H_{(s)}$ in $H_{(-s)}$ is exactly $\{V:V=0\text{ in }\mathbb R^n_+\}$. A continuous antilinear functional on the closed subspace $\dot H_{(s)}$ extends with the same norm to $H_{(s)}$ (orthogonal projection) and is then represented by some $V$; two representatives differ by an element of the annihilator. So the antidual of $\dot H_{(s)}$ is $H_{(-s)}$ modulo the annihilator, that is $\overline H_{(-s)}$, and the norms agree (the infimum over the coset is at most the norm of the norm-preserving extension). Conversely, a functional on the quotient $\overline H_{(-s)}$ is a functional on $H_{(-s)}$ vanishing on the annihilator; it is represented by $u\in H_{(s)}$ orthogonal to the annihilator, and the double annihilator of the closed subspace $\dot H_{(s)}$ is itself. The last statement is Plancherel.

(c) The identity is Plancherel with $(1+|\xi|^2)^k=\sum_{|\alpha|\leq k}\frac{k!}{\alpha!(k-|\alpha|)!}\xi^{2\alpha}$. For (10.1), any extension $U$ gives $N_k(u)^2\leq\sum_{|\alpha|\leq k}\|D^\alpha U\|^2_{L^2(\mathbb R^n)}\leq C\|U\|^2_{(k)}$. For the other inequality, let $c_1,\ldots,c_{k+1}$ solve the Vandermonde system $\sum_{l=1}^{k+1}c_l(-l)^i=1$, $i=0,\ldots,k$ (the nodes $-1,\ldots,-(k+1)$ are distinct). For $u\in\overline{\mathcal S}(\mathbb R^n_+)$ let $Eu=u$ on $x_n\geq0$ and $Eu(x)=\sum_lc_lu(x',-lx_n)$ for $x_n<0$. The normal derivatives of order $i\leq k$ from both sides agree on $x_n=0$, so $Eu\in C^k$, and $\|D^\alpha Eu\|_{L^2(\mathbb R^n_-)}\leq\sum_l|c_l|l^{\alpha_n-1/2}\|D^\alpha u\|_{L^2(\mathbb R^n_+)}$. Hence $\|u\|_{\overline H_{(k)}}\leq\|Eu\|_{(k)}\leq CN_k(u)$ on the dense set $\overline{\mathcal S}(\mathbb R^n_+)$, and by continuity of both sides everywhere. $\square$

Sobolev continuity at integer orders

Theorem 10.3 (Integer orders). Let $a\in S^0_{\mathrm{la}}$ and $k\in\mathbb Z$. Then $T_a$ is bounded on $\dot H_{(k)}(\overline{\mathbb R}{}^n_+)$ and on $\overline H_{(k)}(\mathbb R^n_+)$. These bounded operators are the restrictions of the maps of Theorem 9.1.

Proof. Iterating the commutator identities (5.1) gives, for every $\alpha$,

each $c_{\alpha\beta}$ being a constant-coefficient combination of $x$- and $\xi_n$-derivatives of $a$, linear in $a$. (Indeed $D_jT_c=T_cD_j-iT_{\partial_{x_j}c}-i\delta_{jn}T_{\partial_{\xi_n}c}D_n$, and $\partial_{\xi_n}c\in S^{-1}_{\mathrm{la}}\subset S^0_{\mathrm{la}}$.)

Nonnegative $k$, supported spaces. For $u\in\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)$, $T_au\in\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)$ (Theorem 5.1(d)), and its derivatives on $\mathbb R^n$ are the zero extensions of the derivatives in $\mathbb R^n_+$. By Proposition 10.2(c), (10.2) and Theorem 10.1, $\|T_au\|_{(k)}^2\leq C\sum_{|\alpha|\leq k}\|D^\alpha T_au\|^2_{L^2(\mathbb R^n_+)}\leq C'\sum_{|\beta|\leq k}\|D^\beta u\|^2_{L^2}\leq C''\|u\|^2_{(k)}$. By density (Proposition 10.2(a)) $T_a$ extends to $\dot H_{(k)}$.

Nonnegative $k$, restricted spaces. For $u\in\overline{\mathcal S}(\mathbb R^n_+)$, (10.1), (10.2) and Theorem 10.1 give $\|T_au\|_{\overline H_{(k)}}\leq CN_k(T_au)\leq C'N_k(u)\leq C''\|u\|_{\overline H_{(k)}}$; then use density.

Negative $k$. Let $k\geq0$. For $u\in\overline{\mathcal S}(\mathbb R^n_+)$ and $v\in\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)$, Theorem 7.3 gives $(T_au,v)=(u,T_{a^\dagger}v)$, so $|(T_au,v)|\leq\|u\|_{\overline H_{(-k)}}\|T_{a^\dagger}v\|_{(k)}\leq C\|u\|_{\overline H_{(-k)}}\|v\|_{(k)}$ by the supported case for $a^\dagger\in S^0_{\mathrm{la}}$. By Proposition 10.2(a),(b), $\|T_au\|_{\overline H_{(-k)}}\leq C\|u\|_{\overline H_{(-k)}}$, and density extends $T_a$. In the same way, for $u\in\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)$ and $v\in\overline{\mathcal S}(\mathbb R^n_+)$, $|(T_au,v)|\leq\|u\|_{(-k)}\|T_{a^\dagger}v\|_{\overline H_{(k)}}\leq C\|u\|_{(-k)}\|v\|_{\overline H_{(k)}}$, which bounds $T_a$ on $\dot H_{(-k)}$.

Consistency. The maps of Theorem 9.1 are weakly continuous, the spaces here embed continuously into $\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)$ or $\overline{\mathcal S'}(\mathbb R^n_+)$, and the two definitions agree on the dense subspaces used above. $\square$

All real orders and all Besov exponents

Lemma 10.4 (A mollifier supported in the half space). Let $\phi\in C_0^\infty(\mathbb R^n_+)$ with $\int\phi=1$, and put $\psi=2\phi-\phi*\phi$. Then $\psi\in C_0^\infty(\mathbb R^n_+)$, $\widehat\psi=1-(1-\widehat\phi)^2$, and for all $\zeta\in\mathbb R^n$

With $\psi_\varepsilon(x)=\varepsilon^{-n}\psi(x/\varepsilon)$ and $|s|\leq\tfrac12$,

Proof. $\operatorname{supp}(\phi*\phi)\subset\operatorname{supp}\phi+\operatorname{supp}\phi\subset\mathbb R^n_+$. $\widehat\psi$ is a Schwartz function, and $|1-\widehat\phi(\zeta)|\leq C\min(1,|\zeta|)$ since $\widehat\phi(0)=1$; this gives (10.3). For (10.4) with $|\xi|\leq1$: the first integral is at most $C\int_0^1\varepsilon^{1-2s}d\varepsilon<\infty$ (as $1-2s\geq0$), and the second at most $C|\xi|^4\int_0^1\varepsilon^{1-2s}d\varepsilon$. For $|\xi|\geq1$ substitute $u=\varepsilon|\xi|$ and extend to $(0,\infty)$: the integrals become $|\xi|^{2s-2}\int_0^\infty\min(1,u^{-4})u^{1-2s}du$ and $|\xi|^{2s+2}\int_0^\infty\min(1,u^4)u^{-3-2s}du$, which converge because $1-2s>-1$, $-3-2s<-1$ and $-3+4-2s>-1$. $\square$

The quadratic vanishing of $1-\widehat\psi$ at 0 is needed for $s\geq0$, and it cannot be had with $\psi\geq0$: see Example 10.5. That is why $\psi$ is built from $\phi$ in this way.

Example 10.5 (A positive mollifier is not good enough). Let $0\leq\phi\in C_0^\infty(\mathbb R^n_+)$ with $\int\phi=1$. Its first moment $m=\int x\phi\,dx$ has $m_n>0$, and $\widehat\phi(\zeta)=1-i\zeta\cdot m+O(|\zeta|^2)$. For $\xi=\lambda e_n$ we get $|1-\widehat\phi(\varepsilon\xi)|\geq m_n\varepsilon\lambda/2$ when $\varepsilon\lambda$ is small, so $\int_0^1|1-\widehat\phi(\varepsilon\xi)|^2\varepsilon^{-3-2s}d\varepsilon=\infty$ for $s\geq0$. So the second inequality of (10.4) fails for $\phi$, and it fails for every $\psi\geq0$ supported in $\mathbb R^n_+$: quadratic vanishing forces $\int x_n\psi=0$, which is impossible when $\psi\geq0$ and $x_n>0$ on the support. The function $\psi=2\phi-\phi*\phi$ of Lemma 10.4 takes negative values.

Theorem 10.6 (Sobolev and Besov continuity). Let $a\in S^0_{\mathrm{la}}$, $\sigma\in\mathbb R$ and $1\leq p\leq\infty$. Then $T_a$ is bounded on $\dot B^\sigma_{2,p}(\overline{\mathbb R}{}^n_+)$. In particular:

1. ($p=2$) $T_a$ is bounded on $\dot H_{(\sigma)}(\overline{\mathbb R}{}^n_+)$, and, by duality with $a^\dagger$ (Proposition 10.2(b)), on $\overline H_{(\sigma)}(\mathbb R^n_+)$, for every real $\sigma$.
2. ($p=\infty$) $T_a$ is bounded on $\dot B^\sigma_{2,\infty}(\overline{\mathbb R}{}^n_+)$.

First proof, for $p=2$ (continuous interpolation). Write $\sigma=k+s$ with $k\in\mathbb Z$, $|s|\leq\tfrac12$, and let $u\in\dot H_{(\sigma)}(\overline{\mathbb R}{}^n_+)$. The pieces $\psi_\varepsilon*u$ and $u-\psi_\varepsilon*u$ are supported in $\overline{\mathbb R}{}^n_+$, because $\operatorname{supp}\psi\subset\mathbb R^n_+$; they lie in $\dot H_{(k+1)}$ and $\dot H_{(k-1)}$. By (10.4) and Fubini,

Put $v_\varepsilon=T_a(\psi_\varepsilon*u)$, $w_\varepsilon=T_a(u-\psi_\varepsilon*u)$ and $U=T_au=v_\varepsilon+w_\varepsilon$. By Theorem 10.3 the same integral with $v_\varepsilon,w_\varepsilon$ in place of the two pieces is at most $C'\|u\|^2_{(\sigma)}$. The distribution $U$ already belongs to $H_{(k-1)}$: the input does by the continuous embedding and the compatible integer operator acts there. Thus its Fourier transform is an actual locally square-integrable function. For each $\varepsilon$, the Fourier identity $U=v_\varepsilon+w_\varepsilon$ holds almost everywhere; the integrals below have jointly measurable Fourier representatives. Indeed on any compact interval of positive $\varepsilon$, the Schwartz multiplier estimates and dominated convergence make $\varepsilon\mapsto\psi_\varepsilon*u$ norm continuous in $H_{(k+1)}$, and its complement norm continuous in $H_{(k-1)}$. The bounded integer operators preserve these continuities. Uniform step approximations on that interval therefore converge in the corresponding weighted product $L^2$ spaces. Choose a subsequence whose successive $L^2$ errors are summable; on each bounded frequency set, Cauchy--Schwarz and Fubini make the sum of their absolute differences finite almost everywhere. Its measurable limit represents the norm-continuous family. Exhausting the positive scale intervals and frequency balls gives the required joint representatives and justifies every use of Fubini below. If $\tfrac12<\varepsilon\langle\xi\rangle<1$, then $|\widehat U(\xi)|^2\leq C\big(|\widehat{v_\varepsilon}(\xi)|^2(\varepsilon\langle\xi\rangle)^{2k+2}+|\widehat{w_\varepsilon}(\xi)|^2(\varepsilon\langle\xi\rangle)^{2k-2}\big)$. Multiply by $\varepsilon^{-1-2\sigma}$ and integrate over these $\varepsilon$ (all in $(0,1]$): the left side becomes $c_\sigma\langle\xi\rangle^{2\sigma}|\widehat U(\xi)|^2$ with $c_\sigma=\int_{1/2}^1u^{-1-2\sigma}du>0$, and the right side is at most the integrand of the previous display, evaluated for $v_\varepsilon,w_\varepsilon$ at the frequency $\xi$. Integrating in $\xi$ gives $\|U\|^2_{(\sigma)}\leq C''\|u\|^2_{(\sigma)}$. $\square$

Second proof, for all $p$ (dyadic form). Let $\sigma=k+s$ as before and $u\in\dot B^\sigma_{2,p}(\overline{\mathbb R}{}^n_+)$. Since $B^\sigma_{2,p}\subset H_{(\sigma-\delta)}$ for every $\delta>0$ (the squares $2^{2j(\sigma-\delta)}\|\Pi_ju\|^2_{L^2}$ are at most $2^{-2j\delta}$ times the square of the norm in $B^\sigma_{2,p}$, so they are summable) and $s>-1$, $u\in\dot H_{(k-1)}$. For each $j\geq0$ put $\varepsilon_j=2^{-j}$, $v_j=\psi_{\varepsilon_j}*u\in\dot H_{(k+1)}$, $w_j=u-v_j\in\dot H_{(k-1)}$. Then $T_au=T_av_j+T_aw_j$ (the maps of Theorems 9.1 and 10.3 agree), and, since $\langle\xi\rangle$ is comparable to $2^j$ on $A_j$,

On $A_l$, $|\widehat\psi(2^{-j}\xi)|\leq C\min(1,2^{2(j-l)})$ and $|1-\widehat\psi(2^{-j}\xi)|\leq C\min(1,2^{2(l-j)})$ by (10.3). With $y_l=2^{l\sigma}\|\Pi_lu\|_{L^2}$ this gives

For $|s|\leq\tfrac12$ both brackets are at most $2^{-|j-l|/2}$: for $j\geq l$ they are $2^{(j-l)(s-1)}$ and $2^{(j-l)(s-1)}$; for $j<l$ they are $2^{(j-l)(s+1)}$ and $2^{(j-l)(s+1)}$. Since $(\sum_lc_ly_l^2)^{1/2}\leq\sum_lc_l^{1/2}y_l$, we get $x_j:=2^{j\sigma}\|\Pi_jT_au\|_{L^2}\leq C\sum_l2^{-|j-l|/2}y_l$. Convolution with the summable sequence $2^{-|m|/2}$ is bounded on $\ell^p(\mathbb Z)$ for every $1\leq p\leq\infty$, by the triangle inequality for translates (Section 1), so $\|T_au\|_{B^\sigma_{2,p}}\leq C\|u\|_{B^\sigma_{2,p}}$. Finally $T_au$ is supported in $\overline{\mathbb R}{}^n_+$. $\square$

The dyadic proof treats all $1\leq p\leq\infty$ at once and contains the case $p=2$.

11. Conormal distributions are preserved

Let $\mathcal P_b$ be the set of operators $P=\sum_{|\alpha|\leq M}c_\alpha(x)x_n^{\alpha_n}D^\alpha$ with $c_\alpha\in C^\infty_b(\mathbb R^n)$ and any $M$. For $\kappa\in\mathbb R$ put

the distributions supported in the closed half space that are conormal to the boundary uniformly at infinity.

Lemma 11.1 (Exact compositions). Let $a\in S^m_{\mathrm{la}}$.

(a) If $P\in\mathcal P_b$ has order $\leq M$, then $PT_a=T_{p\star a}$ on $\overline{\mathcal S}(\mathbb R^n_+)$, where

with $q_k$ from (2.1).

(b) Let $M\geq0$ be even and $Q(\xi)=|\xi|^M$. Then $T_Q:=\sum_{|\beta|=M/2}\frac{(M/2)!}{\beta!}D'^{2\beta'}x_n^{2\beta_n}D_n^{2\beta_n}\in\mathcal P_b$ is the operator with compressed symbol $Q^\flat=(|\xi'|^2+x_n^2\xi_n^2)^{M/2}$, and for $f\in S^\mu_{\mathrm{la}}$

(c) If $c\in S^\mu_{\mathrm{la}}$ and $M\geq0$ is an even integer with $M>\mu$, then $c=f\circ Q+g$ with $f,g\in S^0_{\mathrm{la}}$.

Proof. (a) By (5.3), $D_{x_j}(e^{ix\cdot\xi}c^\flat)=e^{ix\cdot\xi}((\xi_j+D_{x_j})c)^\flat$ for $j<n$, and $x_nD_{x_n}(e^{ix\cdot\xi}c^\flat)=e^{ix\cdot\xi}((\xi_n+x_nD_{x_n}+\xi_nD_{\xi_n})c)^\flat$, because $x_n\xi_nc^\flat=(\xi_nc)^\flat$. So $D_jT_c=T_{(\xi_j+D_{x_j})c}$ and $x_nD_nT_c=T_{(\xi_n+x_nD_{x_n}+\xi_nD_{\xi_n})c}$, and $c_\alpha(x)T_c=T_{c_\alpha c}$. These symbol operators preserve lacunarity and raise the order by at most one (the factor $x_n$ is absorbed by the decay in $x_n$). By (2.1), $x_n^{\alpha_n}D^\alpha=D'^{\alpha'}q_{\alpha_n}(x_nD_n)$, which gives (11.2).