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

Boundary operators of every nonnegative real order

This companion retains the complete Section 14 proof of AN03-U032, Totally characteristic operators on the half space. Original author: Claude Opus 5.5 (Anthropic), September 2026; editorial additions: Codex, September 2026; both CC0. Exact prerequisite connections and the analytic justifications below: AN-04 course-writing task and OpenAI Codex, 5 October 2026, CC0. The source's seventeen numbered mathematical displays remain unchanged. Its original illustration is retained with the final output label corrected: $v$ is an operator output, while $F_{f,g}(m)$ is the scalar pairing $(v,g)$.

The approved mathematical antecedent is Hörmander III, 2007 eBook, ISBN 978-3-540-49938-1, Section 18.3. Reading, proof construction and ordinary citation are valid. The complete proof uses the exact programme results linked below; the book citation is not a proof substitute.

Exact earlier proofs

Original section numbers below refer to the four components of the local boundary calculus. Its kernel component proves the full residual bounds; its operator component proves complete composition, adjoints and actual distribution actions; its continuity component proves the order-zero estimates for every real index, full quotient duality, homogeneous finite-seminorm bounds and the residual no-gain obstruction.

The half-space companion H4 proves the right-half-plane logarithm, real one-sided Laplace kernel and exact supported isometries. The Fourier proof and measure proof supply inversion, Plancherel, Fubini and dominated differentiation. The elementary exponential and real integration proofs are in the earlier stationary-phase foundations. Each earlier component retains its own licence.

Analytic facts used in the strip argument

All scalar holomorphic functions below are constructed with continuous real derivatives of every order and the Cauchy--Riemann equations. No regularity theorem for initially nonsmooth complex differentiable functions is needed. On the right half-plane use

The direct derivatives are $L_h=(h+i\tau)^{-1}$ and $L_\tau=i(h+i\tau)^{-1}$. The real logarithm, trigonometric functions and inverse tangent here are those proved in the earlier elementary foundations; the same branch and its derivative are proved in H4. Thus $L$ is smooth and satisfies Cauchy--Riemann, and $\exp(zL)$ is entire in $z$. Its parameter derivatives are $L^j\exp(zL)$.

Here is the exact rectangle maximum estimate needed below. If a complex function $G=u+iv$ is $C^2$ on a neighborhood of a closed rectangle and satisfies Cauchy--Riemann in it, differentiating those equations gives $\Delta u=\Delta v=0$, and hence

For $\delta>0$, the real function $|G|^2+\delta(x^2+y^2)$ has strictly positive Laplacian. Its maximum on the compact rectangle cannot occur in the interior: the second derivative test along the two coordinate lines would give a nonpositive Laplacian there. The maximum is therefore on the boundary. Its boundary value is at most $\sup_{\partial R}|G|^2+\delta\sup_{\partial R}(x^2+y^2)$. Letting $\delta\downarrow0$ gives $\sup_R|G|\le\sup_{\partial R}|G|$. This proves the precise maximum principle used for the smooth scalar pairing in (PS13). Products with the displayed entire exponential normalizations satisfy the same hypotheses.

14. Positive order on both original Sobolev spaces

The following proof extends the zero-order bounds without changing the compressed operator or its distribution action.

The original theorem and spaces

Let $m\ge0$, $s\in\mathbb R$, and let $a\in S^m_{\mathrm{la}}$ take values in $L(\mathbb C^p,\mathbb C^q)$, with $p,q$ fixed finite positive integers. Use the original symbol estimates

The forward Fourier kernel is $e^{-ix\cdot\xi}$, $D=-i\partial$, and

The full-space norm is $\|u\|_{(s)}^2=(2\pi)^{-n}\int(1+|\xi|^2)^s|\widehat u(\xi)|^2\,d\xi$. The supported space is the closed subspace of this $H_{(s)}$ consisting of distributions supported in $\{x_n\ge0\}$. The restricted space is its original ambient restriction quotient, with the infimum norm over $H_{(s)}$ extensions. Neither space is replaced.

We prove

continuously. For each fixed $m,s$, a finite sum $p_{m,s}(a)$ of the original symbol seminorms bounds both operator norms. The maps agree with the original distribution action in Theorem 9.1.

Exact integer-order decomposition

Fix the same normal convolution $\rho$ as Lemma 4.4, including its inverse Fourier convention, $\widehat\rho=1$ near zero and $\operatorname{supp}\widehat\rho\subset(-1/2,1)$. For an integer $M\ge1$ and $a\in S^M_{\mathrm{la}}$, put

Every $w_j\in S^{M-1}_+$. Lemma 4.4 gives $a_j\in S^{M-1}_{\mathrm{la}}$ and $w_j-a_j\in S^{-\infty}_+$, continuously with all the stipulated seminorms. Thus $a_0\in S^{M-2}_+\subset S^{M-1}_+$. It is lacunary because the first definition in (PS4) is a difference of lacunary symbols. The residual sum in the second definition is retained exactly; its separate summands need not themselves be lacunary.

Multiplication of a symbol by $x_n$ preserves every original order and lacunarity: a weighted seminorm uses one extra $x_n$-decay seminorm, and differentiating $x_n$ produces only a derivative of $a_j$. On the original test spaces the exact left-quantization identity is

In the normal term the uncompressed derivative frequency is $\xi_n$; the left factor $x_n$ supplies precisely the original compression $x_n\xi_n$. No derivative is commuted past this factor.

The zero-order theorem is the induction base, for every real $s$, on both spaces. Suppose the bound of order $M-1$ has been proved for every $s$. Each ambient $D_j:H_{(s)}\to H_{(s-1)}$ has norm at most one, since $|\xi_j|^2\le1+|\xi|^2$. Derivatives preserve supported distributions and descend continuously to the restriction quotient. Hence every derivative term in (PS5) maps index $s$ to $s-M$. The term $T_{a_0}$ maps index $s$ to $s-(M-1)$, which embeds into index $s-M$ with norm at most one. The same inequality descends to quotient norms. This proves both integer bounds, with finite original symbol seminorm control.

The identities initially hold on supported or restricted Schwartz test functions. Proposition 10.2(a) gives density for every real index, and Theorem 9.1 gives their weakly continuous distribution actions. Consequently the bounded extensions retain precisely (PS5) and the original distribution action; no boundary delta term is removed by an arbitrary extension. The supported proof uses functions smooth and flat on the boundary before taking density.

A support-preserving exact change of Sobolev index

Set

In dimension one the first term is $1$, with its zero-dimensional Fourier factor. The full factor $\ell$ is retained. In particular

It follows by the original Plancherel norm that

For real $z$ this is equality. The inverse is the literal multiplier $\Lambda_+^{-z}$, so the real-index map is an isometry onto the full $H_{(r-z)}$, and it will be an isometry of the supported subspaces once support is proved. All multipliers have smooth derivatives of polynomial growth; they act continuously on $\mathcal S,\mathcal S'$. On each bounded real strip their Schwartz operator seminorms grow at most as a polynomial in $|\operatorname{Im}z|$ times $e^{\pi|\operatorname{Im}z|/2}$. This follows by differentiating the full $\ell^z$: every derivative is a finite sum of derivatives of $\ell$, powers of $\ell^{-1}$, and polynomial factors in $z$. Parameter derivatives add powers of $\log\ell$, bounded by any fixed positive power of $1+|\xi|$, which proves holomorphy on the test spaces.

Here is a proof of support, with the transform constants. For $\operatorname{Re}q>0$, the elementary Laplace identity gives

We prove this identity for every complex $q$ with $\operatorname{Re}q>0$, without leaving an identity theorem as a prerequisite. Write $\ell=h+i\tau$, $h>0$, and $I(\tau)=\int_0^\infty t^{q-1}e^{-(h+i\tau)t}\,dt$, where $t^{q-1}=\exp((q-1)\log t)$. The absolute near-zero bound is $t^{\operatorname{Re}q-1}$, and the exponential dominates every required large-$t$ factor. Dominated differentiation and integration by parts of $t^qe^{-(h+i\tau)t}$, whose two boundary values vanish, give

The derivative in (PB1) shows that $(h+i\tau)^q I(\tau)$ has derivative zero on the real line. Its value at zero is $\Gamma(q)$. Thus $I(\tau)=\Gamma(q)(h+i\tau)^{-q}$ with precisely the indicated branch. The nonzero denominator is proved next. The same dominated bounds with powers of $\log t$ justify every complex-$q$ derivative on compact subsets of $\operatorname{Re}q>0$.

For clarity, the gamma denominator here is nonzero. Integration by parts in the beta integral gives

Multiply by $N^q$ and put $t=Nu$. The integrand is bounded in modulus by $t^{\operatorname{Re}q-1}e^{-t}$ on $0<t<N$, so dominated convergence gives $\Gamma(q)$. Rewriting the finite product, its limit is

For the real limit, $H_N-\log N$ is positive by comparison with $\int_1^N dx/x$, and is decreasing because $\log(1+1/N)>1/(N+1)$. Hence it converges by completeness of the real numbers. For the complex series, the derivative of the right half-plane logarithm gives

For $|w|\le1/2$ its modulus is at most $|w|^2$. Apply this with $w=q/k$ for all sufficiently large $k$, bounding the finite initial part separately. Thus the displayed series is absolutely convergent, with its actual $O(k^{-2})$ tail. Each logarithm is in the right half-plane branch, so the displayed value is nonzero. This proves the division in (PS9), including complex $q$.

Let

In dimension one $E_t=e^{-t}$. The inverse Fourier kernel of (PS9) is

This is a tempered distribution: near zero the tested integrand has the integrable bound $C t^{\operatorname{Re}q-1}$, and at infinity the factor $e^{-t}$, with finitely many test seminorms, makes it integrable. The kernel is supported in $x_n\ge0$. The tangential Fourier factor is exactly $(2\pi)^{-(n-1)}$; inverse transformation of $e^{-it\xi_n}$ is the displayed delta, without an extra factor.

For arbitrary $z\in\mathbb C$, choose an integer $k\ge0$ with $k>\operatorname{Re}z$. Then $\ell^z=\ell^k\ell^{-(k-z)}$. The second multiplier has the kernel (PS10); the first is the finite operator $(\sqrt{1+|D'|^2}+\partial_n)^k$. Tangential multipliers and normal derivatives preserve normal support. Thus $\Lambda_+^z$ preserves support for every $z$. More explicitly, its action on a Schwartz function supported in the positive half space has that support by (PS10) and the finite operator; the one-sided mollification and compact cutoff approximation in Theorem 9.1(e) extends this conclusion to every supported tempered distribution. All the operators are continuous on $\mathcal S'$, so the support is retained in the weak limit. The same argument applies to $-z$.

It follows that, for real $r$,

are inverse isometries, with exactly the original full-space norms. This proves the support-preserving index change rather than assuming that the multiplier $(1+|D|^2)^{r/2}$ preserves half-space support.

The original analytic symbol family and the strip estimate

Fix $0<m<M$, with $M$ an integer, and retain

The residual contribution is present for every $z$; $A_m=a$ exactly. For each $z$, $A_z\in S^{\operatorname{Re}z}_{\mathrm{la}}$. Frequency differentiation of the displayed scalar factor yields finite polynomials in $z-m$, the full powers of $1+|\xi|^2$, and the original frequency coordinates. Hence each indicated symbol seminorm is bounded by $C(1+|\operatorname{Im}z|)^L p(a)$, uniformly on this real strip, with a finite original seminorm $p$. The convolution and residual bounds of Lemma 4.4 retain the same property. Holomorphy holds in any fixed slightly larger symbol order, such as $S^{M+1}_+$: parameter derivatives produce powers of $\log(1+|\xi|^2)$, which the one extra order bounds. No assertion of order-$M$ holomorphy at its borderline is needed.

For $f,g$ in the original supported Schwartz spaces of dimensions $p,q$, respectively, set

The expression acts in the supported Schwartz space at every stage. The multipliers preserve that space by their full Schwartz estimates and the kernel support proof. The local Theorem 5.1 gives all boundary jets of $T_{A_z}u$; all are zero when the input is supported Schwartz and hence flat on the boundary. Extension by zero is therefore Schwartz, with every seminorm bounded by that theorem. These bounds are continuous in $A_z$ in a fixed sufficiently large symbol order, including the extra order for each prescribed finite number of parameter derivatives. Difference quotients converge by the exponential Taylor formula and those same weighted estimates; the product rule proves complex differentiability of the composition and of its scalar pairing. All real derivatives are continuous. This is holomorphic on the strip, continuous on its closed boundary, and $C^2$ on a neighborhood of each finite closed rectangle. The preceding test-space estimates and Theorem 5.1 bound its growth throughout the strip by $C_{f,g}(1+|\operatorname{Im}z|)^L e^{\pi|\operatorname{Im}z|/2}$. On the two boundary lines, the proved integer bounds, (PS8) and (PS11) give

Take a single finite seminorm $p$ large enough for both endpoints. The zero-order norm depends continuously on finitely many original symbol seminorms, as proved in Section 10. Rescaling a symbol by their sum gives a linear bound in that sum; the integer induction uses only continuous linear symbol operations. Thus the stated $p(a)$ bounds are homogeneous, including $p(a)=0$.

For any fixed $\varepsilon>0$, multiply $F$ by $\exp(\varepsilon(z-m)^2)$. Its boundary bounds now have finite constants $C'_0,C'_M$, since $(1+|\tau|)^L e^{\pi|\tau|/2-\varepsilon\tau^2}$ is bounded. Its modulus tends to zero on the horizontal edges of large rectangles in the strip. Dividing it by $p(a)\|f\|_2\|g\|_2 C'_0$ and multiplying by $\exp[-(z/M)\log(C'_M/C'_0)]$ makes the two vertical-edge bounds at most one. The proved rectangle maximum principle (PB2), followed by the expanding limit, therefore gives

The zero-seminorm or zero-test-function case is immediate and does not require dividing by zero. This is the required strip estimate with its growth control proved.

Since $A_m=a$, duality in the supported $L^2$ space and density of its Schwartz subspace show that $\Lambda_+^{s-m}T_a\Lambda_+^{-s}$ is bounded on supported $L^2$. Conjugating by the exact isometries (PS11) gives the supported bound (PS3) for this real $m$, with finite original seminorm control. Together with the integer case it covers every $m\ge0$ and every real $s$. The bounded extension agrees with the original supported distribution action by test-space density and its weak continuity.

The full restriction quotient

Theorem 7.3 gives $a^\dagger\in S^m_{\mathrm{la}}$, continuously and conjugate-linearly, with the original reversed vector dimensions. Apply the supported result at the index $m-s$:

For the supported/restricted duality of Proposition 10.2(b),

The antidual of $\dot H_{(m-s)}$ is exactly $\overline H_{(s-m)}$ with its original quotient norm, so (PS17) proves the restricted bound, without choosing or identifying a supported extension of the restricted input. Density and Theorem 9.1(d) retain the original quotient distribution action. This completes both assertions of (PS3).

For $m<0$, the same zero-order membership proves boundedness at the same index, but the argument above does not assert a gain of $-m$. Theorem 12.1 gives nonzero residual examples forbidding every such gain. The full residual term, and this exact limitation, remain part of the calculus.