Totally characteristic operators on the half space

Written by Claude Opus 5.5 (Anthropic), September 2026. Self-checked by the writing AI. Public domain (CC0).

Edited and supplemented by Codex, September 2026. The additions and editorial corrections are also public domain (CC0).

This lesson builds the local calculus of totally characteristic pseudodifferential operators on the closed half space $\overline{\mathbb R}{}^n_+=\{x_n\geq0\}$. The differential operators of this kind are generated by the vector fields tangent to the boundary, which are the combinations of $\partial_1,\ldots,\partial_{n-1}$ and $x_n\partial_n$ with smooth coefficients. They respect the boundary: the boundary values of $Pu$ depend only on the boundary values of $u$, and the normal derivatives of $Pu$ of order $k$ at the boundary depend only on those of $u$ of order at most $k$. The pseudodifferential operators of the class come from symbols $a(x,\xi)$ in which the normal frequency $\xi_n$ is replaced by $x_n\xi_n$ before quantization. A condition on the Fourier transform of the symbol in $\xi_n$, called lacunarity, makes the output in the open half space depend only on the input there.

We prove that these operators act on functions and on distributions on the half space, and we compute their commutators and the boundary jets of their outputs. We describe the kernels of the operators of order $-\infty$ near the corner $x_n=y_n=0$ by blowing up the corner, and we show that these kernels are conormal there. The class is closed under adjoints and compositions, with asymptotic formulas for the symbols. Operators of order 0 are bounded on $L^2$, on Sobolev spaces of every real order and on dyadic Besov spaces, and operators of every order preserve the distributions that are conormal to the boundary. In one respect the class differs from the ordinary calculus: operators of order $-\infty$ need not improve regularity at all, because their kernels are singular at the corner.

These operators matter because they are the pseudodifferential counterpart of the differential operators tangent to the boundary. Since they respect the boundary, they can be combined with operators on the boundary, such as taking boundary values. On manifolds with boundary they form the b-calculus, which is used for instance in index theory.

The lesson assumes the symbol calculus on $\mathbb R^n$ from the lesson From symbol estimates to operators on every Sobolev scale; conormal distributions and smooth extension across a boundary from Singularities along a submanifold and smooth boundary passage; and local Besov spaces from Detecting regularity without choosing coordinates. It also uses tempered distributions, the Fourier transform and two facts from functional analysis. Section1 states the precise facts used from the linked lessons and identifies their proof sections.

The freely available author versions [Melrose] and [Loya] provide further reading. The mathematical arguments used here are proved in this lesson and its linked prerequisites.

1. Conventions and background

Notation

Throughout $n\geq1$, $x=(x',x_n)\in\mathbb R^{n-1}\times\mathbb R$, and $$\mathbb R^n_\pm=\{x:\pm x_n>0\},\qquad \overline{\mathbb R}{}^n_+=\{x:x_n\geq0\}.$$ We write $D=-i\partial$, $\widehat u(\xi)=\int e^{-ix\cdot\xi}u(x)\,dx$ (inverse factor $(2\pi)^{-n}$), $\langle\xi\rangle=(1+|\xi|^2)^{1/2}$, and $(u,v)=\int u\overline v$, linear in the first argument. For a function of $(x,\xi)$ we write $a^{(\alpha)}_{(\beta)}=\partial_\xi^\alpha\partial_x^\beta a$.

$C^\infty(\overline{\mathbb R}{}^n_+)$ denotes the functions on $\overline{\mathbb R}{}^n_+$ that are smooth in $\mathbb R^n_+$ and whose derivatives all extend continuously to $\overline{\mathbb R}{}^n_+$. Every such function is the restriction of a smooth function on $\mathbb R^n$; this is the extension fact in the list below. $C^\infty_b$ means smooth with all derivatives bounded.

Sobolev and Besov spaces. $H_{(s)}$ is the space of tempered distributions $u$ whose Fourier transform is locally square integrable and for which the norm $\|u\|_{(s)}^2=(2\pi)^{-n}\int\langle\xi\rangle^{2s}|\widehat u|^2d\xi$ is finite. With the sharp annuli $A_0=\{|\xi|<1\}$, $A_j=\{2^{j-1}\leq|\xi|<2^j\}$ and the Fourier projections $\Pi_j$ onto them, the dyadic Besov norm is $$\|u\|_{B^s_{2,p}}=\big\|\big(2^{js}\|\Pi_ju\|_{L^2}\big)_{j\geq0}\big\|_{\ell^p},\qquad 1\leq p\leq\infty . \tag{1.1}$$ $B^s_{2,p}$ is the space of tempered distributions with locally square-integrable Fourier transform for which this norm is finite. Since $\langle\xi\rangle$ is comparable to $2^j$ on $A_j$, $B^s_{2,2}=H_{(s)}$ with equivalent norms. A distribution $u$ on an open set $\Omega\subset\mathbb R^n$ lies in the local space $B^s_{2,p,\mathrm{loc}}(\Omega)$ if $\chi u$, extended by zero, lies in $B^s_{2,p}$ for every $\chi\in C_0^\infty(\Omega)$.

Values of symbols. All symbols may take values in $L(\mathbb C^p,\mathbb C^q)$ for fixed finite $p,q$. Then $|\cdot|$ is the operator norm, products keep their order, and complex conjugation of a symbol is replaced by the conjugate transpose $a^*$. Every statement below holds in this generality with the same proof, except the square-root step in the proof of Theorem 10.1, where we say what changes. The reader may keep $p=q=1$ in mind.

We also use Peetre’s inequality $(1+|\xi+\zeta|)^s\leq(1+|\xi|)^s(1+|\zeta|)^{|s|}$ for real $s$, which follows from $1+|\xi|\leq(1+|\xi+\zeta|)(1+|\zeta|)$.

Proofs used from earlier lessons

Fourier inversion and Plancherel on $\mathcal S$, $\mathcal S'$ and $L^2$ are proved in the Fourier lesson, Sections1–2 and7–8. The Schwartz kernel theorem for continuous linear maps $\mathcal S(\mathbb R^n)\to\mathcal S'(\mathbb R^n)$ is proved in the Weyl-product lesson, Sections4.1–4.5. Dominated convergence and Fubini’s theorem are proved in the measure chapter, Sections16.1–16.5, and Taylor’s formula in the calculus chapter, Section13.6. The following statements specify the other exact interfaces used below.

Symbols and operators on $\mathbb R^n$. These facts are proved in the lesson From symbol estimates to operators on every Sobolev scale.

Symbol classes. For $m\in\mathbb R$, $S^m(\mathbb R^N\times\mathbb R^k)$ is the space of smooth functions $a(x,\xi)$ with $|\partial_\xi^\alpha\partial_x^\beta a(x,\xi)|\leq C_{\alpha\beta}(1+|\xi|)^{m-|\alpha|}$ for all $x$ and $\xi$. The best constants are seminorms, and they make $S^m$ a Fréchet space. We put $S^{-\infty}=\bigcap_mS^m$. If $a_j\in S^{m_j}$ and $m_j$ decreases to $-\infty$, there is an $a\in S^{m_0}$, unique modulo $S^{-\infty}$, with $a-\sum_{j<l}a_j\in S^{m_l}$ for every $l$; one writes $a\sim\sum_ja_j$. A symbol is polyhomogeneous of degree $\mu$ with step one if $a\sim\sum_ja_j$ with $a_j$ homogeneous of degree $\mu-j$ in $\xi$ for $|\xi|\geq1$.

Quantization and kernels. For $a\in S^m(\mathbb R^n\times\mathbb R^n)$, the operator $\operatorname{Op}(a)u(x)=(2\pi)^{-n}\int e^{ix\cdot\xi}a(x,\xi)\widehat u(\xi)\,d\xi$ maps $\mathcal S$ into $\mathcal S$, and $(a,u)\mapsto\operatorname{Op}(a)u$ is continuous. More generally, let $a$ be any tempered distribution on $\mathbb R^n\times\mathbb R^n$. The formulas $$K_a(x,y)=(2\pi)^{-n}\int e^{i(x-y)\cdot\xi}a(x,\xi)\,d\xi ,\qquad a(x,\xi)=\int e^{-iz\cdot\xi}K_a(x,x-z)\,dz ,$$ in which the integrals are partial Fourier transforms of tempered distributions, are inverse to each other. So every continuous linear map $\mathcal S\to\mathcal S'$ is the map $\operatorname{Op}(a)$ with kernel $K_a$ for exactly one tempered distribution $a$; in particular the operator determines its symbol. If $a$ is a measurable function with polynomial bounds, then $\operatorname{Op}(a)u$ is given, for $u\in\mathcal S$, by the absolutely convergent integral above; one checks this by pairing with a Schwartz function and using Fubini’s theorem and Fourier inversion.

The Gauss transform. $e^{i\langle D_y,D_\eta\rangle}$ is the Fourier multiplier with symbol $e^{i\langle p,q\rangle}$, where $(p,q)$ are the variables dual to $(y,\eta)\in\mathbb R^n\times\mathbb R^n$. It maps $S^m(\mathbb R^n_y\times\mathbb R^n_\eta)$ continuously into itself, and for every $N$ $$e^{i\langle D_y,D_\eta\rangle}c-\sum_{|\alpha|<N}\frac1{\alpha!}\partial_\eta^\alpha D_y^\alpha c\in S^{m-N},$$ each seminorm of the remainder being bounded by finitely many seminorms of $c$. If a sequence is bounded in $S^m$ and converges locally uniformly with all derivatives, then so do its transforms.

The pre-diagonal product estimate. Let $a\in S^{m_1}(\mathbb R^n\times\mathbb R^n)$, $b\in S^{m_2}(\mathbb R^n\times\mathbb R^n)$, and let $B(x,\xi,y,\eta)$ be the transform $e^{i\langle D_y,D_\eta\rangle}$ of $a(x,\eta)b(y,\xi)$ in the variables $(y,\eta)$. Then for all multi-indices and all $N\geq0$, at every point $(x,\xi,y,\eta)$, $$\Big|\partial_\xi^\alpha\partial_x^\beta\partial_\eta^{\alpha'}\partial_y^{\beta'}\Big(B-\sum_{|\gamma|<N}\frac1{\gamma!}\partial_\eta^\gamma a(x,\eta)\,D_y^\gamma b(y,\xi)\Big)\Big|\leq C\langle\eta\rangle^{m_1-N-|\alpha'|}\langle\xi\rangle^{m_2-|\alpha|},$$ with $C$ bounded by finitely many seminorms of $a$ and $b$. At $y=x$, $\eta=\xi$, $B$ is the symbol of $\operatorname{Op}(a)\operatorname{Op}(b)$.

The Schur test. If $K$ is a measurable function on $X\times Y$, for two measure spaces $X$ and $Y$, with $\sup_x\int|K(x,y)|\,dy\leq A$ and $\sup_y\int|K(x,y)|\,dx\leq B$, then $u\mapsto\int K(\cdot,y)u(y)\,dy$ is bounded from $L^2(Y)$ to $L^2(X)$ with norm at most $\sqrt{AB}$.

The preceding unrestricted raw-integral statement is retained for the visible correction and complete proof in Section1.1. Its applications below use Lebesgue output measure.

Sobolev and Besov continuity. For $a\in S^m(\mathbb R^n\times\mathbb R^n)$, $\operatorname{Op}(a)$ maps $H_{(s)}$ continuously into $H_{(s-m)}$ for every real $s$. The spaces $B^s_{2,p}$ are Banach spaces, and multiplication by a $C^\infty_b$ function is bounded on each of them. If $(c_l)_{l\in\mathbb Z}$ is summable, then convolution with it is bounded on $\ell^p(\mathbb Z)$, $1\leq p\leq\infty$, with norm at most $\sum_l|c_l|$, by the triangle inequality for translates.

Local spaces. The local spaces $B^s_{2,p,\mathrm{loc}}$ do not depend on the choice of cutoffs, and they are preserved by multiplication with smooth functions and by changes of coordinates. This is proved in the lesson Detecting regularity without choosing coordinates.

Conormal distributions and extension. These facts are proved in the lesson Singularities along a submanifold and smooth boundary passage.

Conormal distributions. Let $Y$ be a closed submanifold of $\mathbb R^N$ of codimension $k$, and $\mu\in\mathbb R$. The space $I^\mu(\mathbb R^N,Y)$ consists of the $u\in\mathcal D'(\mathbb R^N)$ such that $L_1\cdots L_Mu\in B^{-\mu-N/4}_{2,\infty,\mathrm{loc}}(\mathbb R^N)$ for every $M\geq0$ and all first-order differential operators $L_1,\ldots,L_M$ with smooth coefficients whose principal symbols vanish on the conormal bundle $N^*Y$. Such a $u$ is smooth off $Y$, and its wave front set lies in $N^*Y$. Suppose $Y=\{t=0\}$ in coordinates $x=(t,z)\in\mathbb R^k\times\mathbb R^{N-k}$. Then every $u\in I^\mu(\mathbb R^N,Y)$ with compact support has the form $u(t,z)=\int e^{i\langle t,\tau\rangle}b(z,\tau)\,d\tau$ with $b\in S^{\mu+(N-2k)/4}(\mathbb R^{N-k}\times\mathbb R^k)$ of compact support in $z$, and $b$ is $(2\pi)^{-k}$ times the Fourier transform of $u$ in $t$. Conversely every such integral lies in $I^\mu(\mathbb R^N,Y)$.

Hadamard’s lemma. A smooth function $f(t,z)$ that vanishes on $t=0$ equals $\sum_it_i\int_0^1(\partial_{t_i}f)(st,z)\,ds$. Hence every smooth vector field tangent to $Y=\{t=0\}$ is a combination, with smooth coefficients, of the fields $t_i\partial_{t_j}$ and $\partial_{z_l}$.

Smooth extension. A function on $\overline{\mathbb R}{}^n_+$ that is smooth in $\mathbb R^n_+$, and all of whose derivatives extend continuously to $\overline{\mathbb R}{}^n_+$, is the restriction of a smooth function on $\mathbb R^n$.

Functional analysis.

The Hahn–Banach theorem, seminorm form. Let $p$ be a seminorm on a complex vector space $X$, $M\subset X$ a subspace, and $f$ a linear functional on $M$ with $|f(x)|\leq p(x)$ for $x\in M$. Then $f$ extends to a linear functional $\Lambda$ on $X$ with $|\Lambda(x)|\leq p(x)$ for all $x\in X$. Its full seminorm proof is Section14.7 of the Banach foundations.

Quotients of Fréchet spaces. If $N$ is a closed subspace of a Fréchet space $X$, then $X/N$ with the quotient topology is a Fréchet space. Its full quotient-topology and completeness proof is Section14.8 of the Banach foundations.

1.1. Schur’s bound on the original measure spaces

Editorial correction to the background statement. The preceding Schur statement, as originally written, places no restriction on either measure space. Its assertion about the raw integral is false at that generality. We retain that statement above so the correction is identifiable. Here we prove the counterexample, the bounded operator that exists on the original arbitrary spaces, and its exact relationship with the raw integral. In particular, the raw-integral assertion holds when the output measure is semifinite; the input measure can remain arbitrary. Every application later in this lesson has Lebesgue output measure and therefore satisfies this condition.

Let $(X,\mathcal A,\mu)$ and $(Y,\mathcal B,\nu)$ be arbitrary measure spaces. In this section measurable kernels mean $\mathcal A\otimes\mathcal B$-measurable, finite complex-valued kernels. Retain the original two constants and both pointwise bounds $$\int_Y|K(x,y)|\,d\nu(y)\leq A\quad(x\in X),\qquad \int_X|K(x,y)|\,d\mu(x)\leq B\quad(y\in Y), \qquad 0\leq A,B<\infty . \tag{SC1}$$ All $L^2$ spaces use the original measures and their almost-everywhere equivalence classes; inner products are linear in their first argument. We do not assume completeness, sigma-finiteness or semifiniteness of either measure.

A measurable counterexample with both bounds at every point

Put $I=[0,1]$. On its Borel sets define $$\mu(E)= \begin{cases}0,&E\text{ is meager in }I,\\ \infty,&E\text{ is not meager in }I,\end{cases} \qquad \nu=\text{Lebesgue measure on the Borel sets of }I. \tag{SC2}$$ Here meager means a countable union of sets whose closures have empty relative interior. This is a measure: a countable union of meager sets is meager; if one member of a countable disjoint family is nonmeager, its union is nonmeager and both the sum and union measure are infinite. These two cases also prove countable additivity. The complete-metric Baire theorem, proved in the Banach chapter, Section6, shows that $I$ is not meager in itself. Thus $\mu(I)=\infty$, while every set of finite $\mu$-measure has measure zero.

Enumerate the rationals as $(q_k)_{k\geq1}$, and set $$U_j=\bigcup_{k\geq1} (q_k-2^{-j-k-3},q_k+2^{-j-k-3}),\qquad G=\bigcap_{j\geq1}U_j,\qquad N=\mathbb R\setminus G . \tag{SC3}$$ Every $U_j$ is open and dense, and its Lebesgue measure is at most $$\sum_{k\geq1}2\cdot2^{-j-k-3}=2^{-j-2}. \tag{SC4}$$ Consequently $G$ is a Borel null set. Baire’s theorem on $\mathbb R$ also makes it dense. Each $\mathbb R\setminus U_j$ is closed with empty interior, so $N$ is meager and its complement is Lebesgue null. No finite approximation of these sets is being substituted for them.

Take $X=Y=I$ with the measures in (SC2), and define $$K(x,y)=1_N(x+y),\qquad u(y)=1 . \tag{SC5}$$ Addition is continuous and $N$ is Borel, so $K$ is jointly Borel measurable. For every $x$, the omitted set of $y$’s is $(G-x)\cap I$, a Lebesgue null set; hence its row integral is one. For every $y$, the set of $x$’s where $K\ne0$ is $(N-y)\cap I$. Translation preserves closed sets with empty interior, and restriction of any such set to the nondegenerate interval $I$ has empty relative interior. Thus this section is meager in $I$ and its column integral is zero. We have the exact values $$A=1,\quad B=0,\quad \|u\|_{L^2(\nu)}=1,\qquad \int_I K(x,y)u(y)\,d\nu(y)=1\quad\text{for every }x,\qquad \int_I1^2\,d\mu=\infty . \tag{SC6}$$ The original raw-integral assertion therefore fails. This is not a failure of a matrix estimate or of the square-root constant: the two measures detect different sections of the same measurable set.

The exact $L^2$ operator without measure restrictions

Each $L^2(\nu)$ class has a finite-valued measurable representative $u$. An infinite value, if allowed in a representative, occurs on a null set and can be changed to zero. Its actual nonzero support is sigma-finite: $$S_u=\{u\ne0\}=\bigcup_{k\geq1}\{|u|>1/k\},\qquad \nu\{|u|>1/k\}\leq k^2\|u\|_2^2 . \tag{SC7}$$ The same construction applies to $v\in L^2(\mu)$. On the original restricted measures on $S_v\times S_u$, both factors are sigma-finite. Thus the product and Tonelli/Fubini proofs in the Banach chapter, Sections16.2–16.4 apply there, with no assertion of Tonelli on all of $X\times Y$. Define $$b_K(u,v)= \int_{S_v}\int_{S_u}K(x,y)u(y)\overline{v(x)}\,d\nu(y)\,d\mu(x). \tag{SC8}$$ The two nonnegative integrals needed for product Cauchy–Schwarz obey $$\begin{split} \int_{S_v\times S_u}|K(x,y)|\,|u(y)|^2\,d(\mu\otimes\nu) &\leq B\|u\|_{L^2(\nu)}^2,\\ \int_{S_v\times S_u}|K(x,y)|\,|v(x)|^2\,d(\mu\otimes\nu) &\leq A\|v\|_{L^2(\mu)}^2,\\ \int_{S_v\times S_u}|K(x,y)|\,|u(y)|\,|v(x)|\,d(\mu\otimes\nu) &\leq\sqrt{AB}\,\|u\|_{L^2(\nu)}\|v\|_{L^2(\mu)} . \end{split} \tag{SC9}$$ Here the first line integrates the column bound and the second integrates the row bound. Cauchy–Schwarz uses the two displayed functions $|K|^{1/2}|u|$ and $|K|^{1/2}|v|$, retaining both factors. All integrals on the right are finite, including when $A=0$ or $B=0$; in either zero case the third integral vanishes. These bounds prove absolute convergence of (SC8). Changing $u$ or $v$ on its original null set changes no pairing: on the sigma-finite union of the old and new supports, those changes are product-null by the proved product theorem. The same observation on finite unions of supports proves sesquilinearity.

We include the Hilbert-space facts needed here at this generality. Cauchy–Schwarz for any measure follows by integrating $|f-tg|^2\geq0$ with its minimizing complex $t$, treating $g=0$ separately; it gives Minkowski by expanding $\|f+g\|_2^2$. If a sequence is Cauchy in $L^2(\mu)$, choose a subsequence with successive differences $d_j$ of norms at most $2^{-j}$. For $s_m=\sum_{j\leq m}|d_j|$, Minkowski and monotone convergence give $$\int\left(\sum_{j\geq1}|d_j|\right)^2d\mu =\lim_m\int s_m^2d\mu \leq\left(\sum_{j\geq1}2^{-j}\right)^2 . \tag{SC10}$$ The sum is finite off a measurable null set. There the subsequence converges pointwise; define the limit to be zero on that null set. Applying the same bound to each tail proves convergence in $L^2$, and the Cauchy property gives convergence of the whole sequence. This proves completeness on arbitrary $\mu$, and the proof for $\nu$ is identical.

For completeness, the representing-vector argument requires no separability. For fixed $u$, put $\ell(v)=\overline{b_K(u,v)}$, a bounded linear functional on this Hilbert space. If $\ell=0$, choose its representing vector to be zero. Otherwise the closed affine set $\ell(v)=1$ is nonempty and its distance $d$ from zero is positive, since $1\leq\|\ell\|\|v\|$. A sequence in that set whose norms decrease to $d$ is Cauchy: the parallelogram identity gives $$\|z_j-z_k\|^2 =2\|z_j\|^2+2\|z_k\|^2-4\|(z_j+z_k)/2\|^2 \leq2\|z_j\|^2+2\|z_k\|^2-4d^2 . \tag{SC11}$$ Completeness gives a minimizing vector $z$ in the set. Varying $z$ by $tw$ and $itw$, for real $t$ and $w\in\ker\ell$, shows $\langle z,w\rangle=0$. Since $v-\ell(v)z\in\ker\ell$, $$\ell(v)=\left\langle v,\frac{z}{\|z\|^2}\right\rangle,\qquad b_K(u,v)=\left\langle\frac{z}{\|z\|^2},v\right\rangle . \tag{SC12}$$ The vector in the second pairing is unique, as its difference from another would pair to zero with itself. Define it to be $T_Ku$. Sesquilinearity and uniqueness prove that $T_K$ is linear. Taking $v=T_Ku$ in (SC9), and treating its zero norm separately, proves $$T_K:L^2(\nu)\longrightarrow L^2(\mu),\qquad \langle T_Ku,v\rangle=b_K(u,v),\qquad \|T_K\|\leq\sqrt{AB}. \tag{SC13}$$ This operator is defined on the original spaces. Its adjoint retains the literal swapped kernel $K^*(y,x)=\overline{K(x,y)}$: absolute Fubini on the two supports gives $b_K(u,v)=\overline{b_{K^*}(v,u)}$, so $T_K^*=T_{K^*}$. The swapped constants are $B,A$.

The raw integral on every finite-measure piece

Fix $u$ and its support in (SC7). For any nonnegative product-measurable function $h$ on $X\times S_u$, its section integral is a measurable function of $x$, without a condition on $\mu$. Indeed, on a finite-$\nu$ piece the section-measure argument in Section16.3 of the linked chapter uses rectangles, finite subtraction for complements, and countable sums for disjoint unions; its generating-class proof in Section16.2 requires no outer measure hypothesis. Increasing finite-measure exhaustions of $S_u$, followed by increasing simple approximations to $h$, give the assertion. Real and imaginary parts then give measurability of absolutely convergent complex section integrals.

Consequently the following sets and functions are measurable: $$J_u(x)=\int_{S_u}|K(x,y)|\,|u(y)|\,d\nu(y),\quad D_u=\{J_u=\infty\},\qquad r_0(u)(x)= \begin{cases} \displaystyle\int_YK(x,y)u(y)\,d\nu(y),&x\notin D_u,\\ 0,&x\in D_u . \end{cases} \tag{SC14}$$ The value zero on $D_u$ is an explicit convention; it is not an assertion that the divergent raw integral exists there. For a measurable $E\subset X$ with $\mu(E)<\infty$, Tonelli on $E\times S_u$ and the original column bound give $$\int_E\int_{S_u}|K(x,y)|\,|u(y)|^2d\nu(y)d\mu(x) \leq B\|u\|_2^2 . \tag{SC15}$$ The inner quantity is finite for almost every $x$ in $E$. Weighted Cauchy–Schwarz in $y$ gives, at those points, $$J_u(x)^2 \leq\left(\int_Y|K(x,y)|d\nu(y)\right) \left(\int_{S_u}|K(x,y)|\,|u(y)|^2d\nu(y)\right). \tag{SC16}$$ If $A=0$, every row is zero almost everywhere in $\nu$, and $J_u=0$ at every $x$; this also covers the zero-row case before multiplying any infinite value. Otherwise (SC15)–(SC16) prove $$\mu(E\cap D_u)=0,\qquad \int_E|r_0(u)|^2d\mu\leq AB\|u\|_2^2 \quad\text{for every }E\text{ of finite measure}. \tag{SC17}$$ For $v$ supported in such $E$, absolute Fubini and (SC9) identify $\langle r_0(u),v\rangle=b_K(u,v)$ on $E$. Both $r_0(u)1_E$ and $(T_Ku)1_E$ lie in $L^2(\mu)$; testing their difference proves $$\int_E|r_0(u)-T_Ku|^2d\mu=0 \quad(\mu(E)<\infty). \tag{SC18}$$ Equality on all finite-measure pieces is the exact comparison proved so far. In the counterexample it does not imply equality almost everywhere for the original $\mu$.

The space defined by this defect and its complete quotient map

Let $F_\mu$ be the vector space of original almost-everywhere classes of finite-valued measurable functions $h$ for which $$p_\mu(h)=\sup_{\substack{E\in\mathcal A\\\mu(E)<\infty}} \left(\int_E|h|^2d\mu\right)^{1/2}<\infty,\qquad N_\mu=\{h\in F_\mu:p_\mu(h)=0\}. \tag{SC19}$$ Minkowski on each $E$ proves that $p_\mu$ is a seminorm. Its kernel consists exactly of the functions invisible to every finite-measure test. This definition retains the original global null classes, so $N_\mu$ can contain a class that is nonzero globally.

There is a canonical onto isometry $$i_\mu:L^2(\mu)\longrightarrow F_\mu/N_\mu,\qquad i_\mu(h)=h+N_\mu . \tag{SC20}$$ First, for $h\in L^2(\mu)$, the finite-measure level sets in (SC7), made increasing by finite unions, exhaust its support. Monotone convergence gives $p_\mu(h)=\|h\|_2$. Thus this map is injective and isometric.

To prove surjectivity, take $h\in F_\mu$, and put $c=p_\mu(h)^2$. If $c=0$, its class in the quotient is zero. If $c>0$, choose finite-measure sets $E_k$ such that $\int_{E_k}|h|^2d\mu>c-2^{-k}$. Replace these sets by the finite unions $F_k=\bigcup_{j\leq k}E_j$, and put $S=\bigcup_kF_k$. Then $$\int_S|h|^2d\mu=\lim_k\int_{F_k}|h|^2d\mu=c,\qquad h_s=h1_S\in L^2(\mu). \tag{SC21}$$ For any finite-measure $E$, the disjoint union $F_k\cup(E\setminus S)$ still has finite measure, so $$\int_{F_k}|h|^2d\mu+\int_{E\setminus S}|h|^2d\mu\leq c . \tag{SC22}$$ Letting $k$ increase proves that the second integral is zero. Therefore $h-h_s\in N_\mu$, establishing surjectivity. Two choices of $h_s$ differ by an $L^2$ element of $N_\mu$, hence are equal in the original $L^2$ space. The inverse $P_\mu=i_\mu^{-1}$ is consequently canonical and linear, and the quotient is a complete Hilbert space through this isometry.

Equations (SC17)–(SC18) now give the complete connecting maps $$r_0(u)\in F_\mu,\qquad r_0(u)+N_\mu=i_\mu(T_Ku),\qquad P_\mu(r_0(u)+N_\mu)=T_Ku . \tag{SC23}$$ In particular the raw assignment followed by the quotient is linear, even though the conventions on divergent sets were only pointwise choices. Any other finite measurable convention on $D_u$ has the same quotient class, because every finite-measure intersection of $D_u$ is null. In (SC2), $p_\mu$ is zero on every finite-valued measurable function, $F_\mu=N_\mu$, and $L^2(\mu)=\{0\}$. The raw constant one in (SC6) is a nonzero original class in $N_\mu$; (SC13) is the zero operator. Thus (SC23) describes exactly what the counterexample loses under finite-measure testing.

When the raw-integral assertion is recovered

A measure is semifinite if each measurable set of positive measure contains a measurable subset of finite positive measure. Under this condition, a measurable set whose intersection with every finite-measure set is null is itself null: if it were positive, the defining subset would contradict the intersection property. Applying this first to $D_u$ and then to the sets $\{|h|>1/k\}$ for $h\in N_\mu$ gives $$\mu\text{ semifinite}\quad\Longrightarrow\quad \mu(D_u)=0,\quad N_\mu=\{0\},\quad r_0(u)=T_Ku\ \mu\text{-almost everywhere},\quad \|r_0(u)\|_2\leq\sqrt{AB}\|u\|_2 . \tag{SC24}$$ There is no restriction on the input measure $\nu$ in this implication.

The defect criterion is exact: $N_\mu=\{0\}$ if and only if $\mu$ is semifinite. Indeed, if semifiniteness fails, its definition supplies a positive measurable set $H$ with no finite positive-measure subset. Every finite-measure intersection $E\cap H$ has measure zero. The finite-valued function $1_H$ therefore belongs to $N_\mu$ and is a nonzero original almost-everywhere class. This proves the converse. Equivalently, semifiniteness gives $$\mu(H)=\sup_{\substack{E\subset H\\\mu(E)<\infty}}\mu(E). \tag{SC25}$$ For finite $\mu(H)$ this is immediate. If $\mu(H)=\infty$ and the supremum $c$ were finite, choose finite-measure $E_k\subset H$ approaching $c$, take their increasing finite unions, and let their union be $S$. Measure continuity gives $\mu(S)=c$. Any finite positive-measure subset of $H\setminus S$ would raise the supremum by its disjoint union with $S$, while $H\setminus S$ still has infinite measure. This contradicts semifiniteness and proves (SC25). We have proved the exact vanishing criterion for the defect space; we do not infer that every kernel on a nonsemifinite space must fail its raw-integral bound.

Finally, the same construction works for the finite vector dimensions in this lesson, retaining their order. If $K(x,y)\in L(\mathbb C^p,\mathbb C^q)$, replace $|K|$ in (SC1) and (SC9) by its original operator norm and use $$|\langle K(x,y)u(y),v(x)\rangle_{\mathbb C^q}| \leq\|K(x,y)\|\,|u(y)|_{\mathbb C^p}|v(x)|_{\mathbb C^q}. \tag{SC26}$$ This gives exactly (SC9), with no dimension factor. Vector-valued $L^2$ completeness follows by the same norm-sum proof (SC10); the Hilbert representing argument (SC11)–(SC12) applies unchanged. The raw convergence proof bounds the norm of its vector integral by $J_u$. Define $F_\mu,N_\mu$ with the original Euclidean vector norm; (SC21)–(SC22) prove the same onto isometry and comparison. The adjoint kernel is exactly $K(y,x)^*$, reversing the two vector dimensions, as absolute Fubini in (SC8) shows. All subsequent Schur applications thus retain the full scalar or matrix constants and the original Lebesgue integral.

2. Totally characteristic differential operators

Let $\mathcal V_b$ be the smooth vector fields $V=\sum_jv_j\partial_j$, $v_j\in C^\infty(\overline{\mathbb R}{}^n_+)$, that are tangent to the boundary, that is, $v_n(x',0)=0$. Let $\operatorname{Diff}_b(\overline{\mathbb R}{}^n_+)$ be the algebra of operators on $C^\infty(\overline{\mathbb R}{}^n_+)$ generated by $\mathcal V_b$ and by multiplication with functions in $C^\infty(\overline{\mathbb R}{}^n_+)$, and $\operatorname{Diff}^m_b$ the span of products containing at most $m$ vector fields. Its elements are the totally characteristic differential operators.

Proposition 2.1 (Structure of totally characteristic differential operators).

1. $\mathcal V_b$ is the $C^\infty(\overline{\mathbb R}{}^n_+)$-module generated by $\partial_1,\ldots,\partial_{n-1}$ and $x_n\partial_n$.

2. For every integer $k\geq0$, $$x_n^kD_n^k=\prod_{j=0}^{k-1}\big(x_nD_n+ij\big)=:q_k(x_nD_n). \tag{2.1}$$ Hence $\{x_n^jD_n^j:j\leq k\}$ and $\{(x_nD_n)^j:j\leq k\}$ span the same space, with constant coefficients.

3. $\operatorname{Diff}^m_b$ consists exactly of the finite sums $$P=\sum_{|\alpha|\leq m}c_\alpha(x)\,x_n^{\alpha_n}D^\alpha,\qquad c_\alpha\in C^\infty(\overline{\mathbb R}{}^n_+), \tag{2.2}$$ equivalently of the sums $\sum_{|\alpha|\leq m}c'_\alpha(x)D'^{\alpha'}(x_nD_n)^{\alpha_n}$.

4. For $P$ as in (2.2) and $u\in C^\infty(\overline{\mathbb R}{}^n_+)$, $(Pu)(x',0)=\sum_{\alpha_n=0}c_\alpha(x',0)D'^{\alpha'}u(x',0)$: the boundary value of $Pu$ depends only on the boundary value of $u$.

Proof. (a) If $v_n(x',0)=0$, then $v_n(x)=x_nw(x)$ with $w(x)=\int_0^1(\partial_nv_n)(x',\theta x_n)\,d\theta\in C^\infty(\overline{\mathbb R}{}^n_+)$. Thus $V=\sum_{j<n}v_j\partial_j+w\,x_n\partial_n$. Conversely each generator is tangent.

2. For $k\geq0$ and $u$ smooth, $x_nD_n(x_n^kD_n^ku)=x_n^{k+1}D_n^{k+1}u+x_n(D_nx_n^k)D_n^ku=x_n^{k+1}D_n^{k+1}u-ik\,x_n^kD_n^ku$, because $D_nx_n^k=-ikx_n^{k-1}$. So $x_n^{k+1}D_n^{k+1}=(x_nD_n+ik)\,x_n^kD_n^k$, and induction gives (2.1). The polynomial $q_k$ is monic of degree $k$, so the triangular system can be inverted.

3. Moving a function to the left across a generator produces only multiplication operators: $\partial_jc=c\partial_j+(\partial_jc)$ and $x_n\partial_nc=c\,x_n\partial_n+x_n(\partial_nc)$. So a product of at most $m$ vector fields and functions is a sum of terms $c(x)M_1\cdots M_l$, $l\leq m$, with each $M_i$ one of the generators in (a). These generators commute pairwise, because $[x_n\partial_n,\partial_j]=0$ for $j<n$. So each word is a constant times $D'^{\beta'}(x_nD_n)^{\beta_n}$ with $|\beta|\leq m$, and (b) rewrites it in the form (2.2). Conversely $x_n^{\alpha_n}D^\alpha=D'^{\alpha'}q_{\alpha_n}(x_nD_n)$ is a product of $|\alpha|$ generators.

4. At $x_n=0$ every term with $\alpha_n>0$ carries the factor $x_n^{\alpha_n}$, which vanishes. $\square$

Part (d) is the motivation for the whole lesson. An operator that respects the boundary in this way can be followed by boundary operators. Theorem 5.1(c) below extends (d) to the pseudodifferential operators of this lesson and to normal derivatives of every order.

3. Function spaces on the half space

Restrictions and supports

Two ways to attach a space to the half space. Let $F$ be a space of distributions on $\mathbb R^n$.

• $\overline F(\mathbb R^n_+)$ is the space of restrictions $U|_{\mathbb R^n_+}$, $U\in F$, with the quotient topology of $F/\{U\in F:U=0\text{ in }\mathbb R^n_+\}$.
• $\dot F(\overline{\mathbb R}{}^n_+)$ is the space of $U\in F$ with $\operatorname{supp}U\subset\overline{\mathbb R}{}^n_+$, with the topology of $F$.

These are different objects and must be kept apart. A restriction of a Schwartz function may have any boundary values. A Schwartz function supported in $\overline{\mathbb R}{}^n_+$ vanishes to infinite order on $x_n=0$, since all its derivatives are continuous and vanish for $x_n<0$. The zero extension of an element of $\overline{\mathcal S}(\mathbb R^n_+)$ is an integrable function in $\dot{\mathcal S}'(\overline{\mathbb R}{}^n_+)$; it lies in $\dot{\mathcal S}(\overline{\mathbb R}{}^n_+)$ only when all its normal derivatives vanish at the boundary. In this notation, $C^\infty(\overline{\mathbb R}{}^n_+)=\overline{C^\infty}(\mathbb R^n_+)$, by the smooth extension fact of Section 1.

Lemma 3.1 (Supports in the closed half space). Let $U\in\mathcal S'(\mathbb R^n)$ with $\operatorname{supp}U\subset\overline{\mathbb R}{}^n_+$, and let $\varphi\in\mathcal S(\mathbb R^n)$ vanish in $\mathbb R^n_+$. Then $U(\varphi)=0$.

Proof. First, $U(\psi)=0$ whenever $\psi\in\mathcal S$ vanishes on a neighbourhood $W$ of $\operatorname{supp}U$: for $\psi\in C_0^\infty$ this is the definition of the support, and in general $\psi\,\theta(\cdot/R)\to\psi$ in $\mathcal S$ for a cutoff $\theta$ equal to 1 near 0, while each $\psi\theta(\cdot/R)$ vanishes on $W$. Now put $\varphi_\delta(x)=\varphi(x',x_n+\delta)$. It vanishes on $\{x_n>-\delta\}$, a neighbourhood of $\overline{\mathbb R}{}^n_+$, so $U(\varphi_\delta)=0$; and $\varphi_\delta\to\varphi$ in $\mathcal S$ as $\delta\to0$. $\square$

Restricted Schwartz functions

For $v\in\overline{\mathcal S}(\mathbb R^n_+)$ and multi-indices $\alpha,\beta$ put $$q_{\alpha,\beta}(v)=\sup_{x\in\mathbb R^n_+}|x^\alpha D^\beta v(x)| .$$

Lemma 3.2 (Restricted Schwartz functions).

1. Each $q_{\alpha,\beta}$ is finite and continuous on $\overline{\mathcal S}(\mathbb R^n_+)$. For every continuous seminorm $q$ on $\overline{\mathcal S}(\mathbb R^n_+)$ there are $k$ and $C$ with $$q(v)\leq C\sum_{|\alpha|+|\beta|\leq2k}q_{\alpha,\beta}(v) . \tag{3.1}$$ So the $q_{\alpha,\beta}$ define the quotient topology.

2. $\overline{\mathcal S}(\mathbb R^n_+)$ is a Fréchet space.

3. A function $w\in C^\infty(\overline{\mathbb R}{}^n_+)$ is the restriction of a Schwartz function if and only if every $q_{\alpha,\beta}(w)$ is finite.

4. If $g\in\overline{\mathcal S}(\mathbb R^n_+)$, $k\geq1$, and $\partial_n^jg(x',0)=0$ for $j<k$, then $g=x_n^kh$ with $h\in\overline{\mathcal S}(\mathbb R^n_+)$.

Proof. (a) For every extension $V$ of $v$, $q_{\alpha,\beta}(v)\leq\sup_{\mathbb R^n}|x^\alpha D^\beta V|$; so $q_{\alpha,\beta}$ is bounded by a quotient seminorm, hence finite and continuous. Conversely let $q$ be continuous, and let $\pi$ be the restriction map. Then $q\circ\pi$ is a continuous seminorm on $\mathcal S(\mathbb R^n)$, so there are $k,C$ with $q(\pi V)\leq C\sum_{|\alpha|+|\beta|\leq k}\sup_{\mathbb R^n}|x^\alpha D^\beta V|$. Fix $V\in\mathcal S$. Put $\tilde V=V$ on $x_n\geq0$ and $$\tilde V(x)=\theta(x_n)\sum_{j\leq k}\partial_n^jV(x',0)\frac{x_n^j}{j!}\qquad(x_n<0),$$ with $\theta\in C_0^\infty(\mathbb R)$ equal to 1 on $(-1,1)$. The two pieces have the same derivatives of order $\leq k$ on $x_n=0$, so $\tilde V\in C^k(\mathbb R^n)$. For $|\alpha|+|\beta|\leq k$, $\sup_{\mathbb R^n}(1+|x|)^{|\alpha|}|D^\beta\tilde V|$ is bounded by a constant times $\sum_{|\alpha'|+|\gamma|\leq2k}q_{\alpha',\gamma}(\pi V)$: on $x_n\geq0$ this is clear, and on $x_n<0$ the derivatives are combinations of derivatives of $\theta(x_n)x_n^j$ (bounded, with support in a fixed interval) and of $D'^{\beta'}\partial_n^jV(x',0)$, $|\beta'|+j\leq2k$, whose weighted suprema are limits from $\mathbb R^n_+$. Now take $\phi\in C_0^\infty(\mathbb R^n_-)$ with $\int\phi=1$, $\phi_\varepsilon(x)=\varepsilon^{-n}\phi(x/\varepsilon)$, $0<\varepsilon\leq1$. Then $\tilde V*\phi_\varepsilon\in\mathcal S(\mathbb R^n)$. For $x\in\mathbb R^n_+$ the convolution only uses values at $x-y$ with $(x-y)_n>x_n>0$, so $\pi(\tilde V*\phi_\varepsilon)=\pi(V*\phi_\varepsilon)$. For $|\beta|\leq k$, $D^\beta(\tilde V*\phi_\varepsilon)=(D^\beta\tilde V)*\phi_\varepsilon$, and $1+|x|\leq(1+R)(1+|x-y|)$ for $y\in\operatorname{supp}\phi_\varepsilon\subset\{|y|\leq R\}$. Hence $q(\pi(V*\phi_\varepsilon))\leq C'\sum_{|\alpha|+|\beta|\leq2k}q_{\alpha,\beta}(\pi V)$, uniformly in $\varepsilon$. Since $V*\phi_\varepsilon\to V$ in $\mathcal S$, (3.1) follows.