Cutoffs transported through adaptive neighborhoods

Localization must remain controlled when differentiated along the evolution and Hamiltonian directions. A cutoff supported on a long curved neighborhood can still have bounded transverse derivatives: its shortest transverse width is the scale used to measure those derivatives. The canonical map and the linear frequency term must both respect that scale.

We prove the full cutoff and transport bounds for the two neighborhood families in Covering by curved Hamiltonian neighborhoods. The long type-II family uses the radius $R_2$; the separate type-I family uses $R_1$. The proof also covers every ordered component of the type-II partition.

The original scalar hypotheses, canonical maps, coefficient estimates and residual estimates are established in An adaptive scale for repeated brackets, An explicit canonical cell for a large transverse gradient, and The linear frequency coefficient and the bracket scale.

1. Symbol bounds with time as a parameter

For a radius $r>0$, write

g_r=|dw|^2/r^2. \tag{1.1}

A family belongs uniformly to $S(A,g_r)$, for a positive amplitude $A$ constant in $w$, if for every transverse multiindex $\alpha$,

|D_w^\alpha f(t,w)|\leq C_\alpha A r^{-|\alpha|}. \tag{1.2}

Time is a parameter in (1.2); the constants must be uniform in time. There is no time differential in $g_r$. We use the Poisson convention

\{q,f\}= \sum_{j\geq2}(q_{\xi_j}f_{x_j}-q_{x_j}f_{\xi_j}), \qquad H_qf=\{q,f\}, \tag{1.3}

and let $D_t=-i\partial_t$. Define $\mathcal T=D_t+H_q$. Each estimate below bounds the time derivative and Hamiltonian derivative separately, so it also applies when either convention inserts a fixed complex factor in the Hamiltonian term.

Let $\Phi_j,\psi_j,\Psi_j$ be the type-II cutoffs and ordered components from the cover theorem, with center scales $M_j$. Let $\varphi_a$ be its type-I cutoffs, with center scales $m_a$.

Theorem 1.1 (transported cutoff bounds). For $\rho\geq1$ fixed before sufficiently small $\lambda$, the following families are bounded uniformly in their index, time and $\rho$:

\begin{gathered} \Phi_j,\ \psi_j,\ \Psi_j\in S(1,g_{R_2}),\\ M_j^{-1}\mathcal T\Phi_j\in S(1,g_{R_2}),\\ M_j^{-1}\mathcal T\psi_j\in S(1,g_{R_2}),\\ M_j^{-1}\mathcal T\Psi_j\in S(1,g_{R_2}),\\[3pt] \varphi_a\in S(1,g_{R_1}),\\ m_a^{-1}\mathcal T\varphi_a\in S(1,g_{R_1}). \end{gathered} \tag{1.4}

The permitted upper bound for $\lambda$ can depend on the fixed $\rho$. There is no claim that the type-I cutoffs satisfy the stronger $R_2$ derivative bounds.

2. Canonical composition preserves the transverse symbol class

At a type-II center write $a$ for the selected gradient. The exact canonical map and inverse satisfy, for every positive derivative order,

|D^\gamma\chi|+|D^\gamma\chi^{-1}| \leq C_\gamma a^{1-|\gamma|}. \tag{2.1}

Moreover

a>\rho M/R,\qquad R_2/a\longrightarrow0, \tag{2.2}

by the finite-bracket lower bound and $\kappa+\kappa_2<2/(k+1)$.

Suppose $f$ satisfies $|D^\beta f|\leq C_\beta A R_2^{-|\beta|}$ in the canonical chart. In an order-$|\alpha|$ derivative of $f\circ\chi^{-1}$, a chain-rule term with $r$ derivatives on $f$ and $r$ factors of differentiated inverse maps is bounded by

C_\alpha A R_2^{-r}a^{r-|\alpha|} =C_\alpha A R_2^{-|\alpha|} (R_2/a)^{|\alpha|-r}. \tag{2.3}

Here $1\leq r\leq|\alpha|$; zeroth order is immediate. The finite chain-rule sums and $R_2/a\leq1$ prove preservation of $S(A,g_{R_2})$. The same argument with $\chi$ gives the converse comparison. Compact support inside the chart permits smooth extension by zero without creating additional boundary terms.

The canonical maps are independent of the running time. Thus time differentiation commutes with pullback. Symplectic invariance gives

(\mathcal Tf)\circ\chi =D_t(f\circ\chi)+\{Q,f\circ\chi\}, \qquad Q=q\circ\chi. \tag{2.4}

Translations by the fixed center are understood in (2.4). It is therefore enough to prove all type-II primitive bounds in the canonical chart.

3. Primitive cutoff derivatives

Suppress the center index and write

\begin{gathered} \ell_0=\rho A_2=B_2^{-1},\qquad \ell=\max(\ell_0,R_2)=(B_2')^{-1},\\ L=(\rho M)^{1/2},\qquad b=\min(B_2,L^{-1}). \end{gathered} \tag{3.1}

In particular

b\leq B_2'\leq B_2,\qquad B_2'\leq R_2^{-1},\qquad \rho A_2B_2=1. \tag{3.2}

The first inequality uses $R_2<L$: the reciprocal of $b$ is $\max(B_2^{-1},L)$, while the reciprocal of $B_2'$ is $\max(B_2^{-1},R_2)$.

Both primitive cutoffs in the cover theorem are fixed smooth functions of

M(t-t_c),\quad z/\ell,\quad \xi/R_2,\quad u/R_2, \tag{3.3}

including the smooth squared-norm factor in $u$. Rescaling and the product rule give, for $f=\Phi$ or $\psi$,

\begin{gathered} |D_w^\alpha f|\leq C_\alpha R_2^{-|\alpha|},\\ |D_w^\alpha\partial_z f| \leq C_\alpha B_2'R_2^{-|\alpha|},\\ |D_w^\alpha M^{-1}D_tf| \leq C_\alpha R_2^{-|\alpha|}. \end{gathered} \tag{3.4}

The bound in the second line retains the longer spatial width instead of discarding it. This will cancel the amplitude of the linear frequency coefficient.

4. The residual Hamiltonian term is small

Decompose the transformed symbol as

Q(t,z,\xi,u)=\mathcal R(t,z,\xi,u)+\xi c(t,z), \qquad c(t,z)=\partial_\xi Q(t,z,0,0). \tag{4.1}

The canonical residual estimate gives, for every transverse multiindex $\beta$,

|D_w^\beta\mathcal R| \leq C_\beta\rho M b^{\beta_z}L^{-\beta_\xi-|\beta_u|}. \tag{4.2}

Since $b\leq L^{-1}$, every additional transverse derivative costs at most $L^{-1}$. On the support of either cutoff, all charts lie within the domain of that estimate.

Differentiate a term of $M^{-1}\{\mathcal R,f\}$ by $D_w^\alpha$. Split the additional derivatives into $\beta$ on the residual derivative and $\alpha-\beta$ on the cutoff derivative. Equations (4.2), (3.4) bound it by

\begin{gathered} C_\alpha\rho L^{-1-|\beta|}R_2^{-1-|\alpha-\beta|}\\ =C_\alpha\frac{\rho}{R_2L} R_2^{-|\alpha|}(R_2/L)^{|\beta|}. \end{gathered} \tag{4.3}

There are finitely many Poisson coordinates and product-rule terms. Since $R_2/L<1$, this proves

M^{-1}\{\mathcal R,f\} \in S\bigl(\rho/(R_2L),g_{R_2}\bigr). \tag{4.4}

The amplitude $\rho/(R_2L)$ tends to zero for fixed $\rho$, hence is bounded by one for sufficiently small $\lambda$. This controls every transverse derivative, including those falling on the residual and on the cutoff simultaneously.

5. The linear frequency term uses the long spatial width

The coefficient estimates give, at every nonnegative spatial derivative order,

M^{-1}|\partial_z^r c(t,z)| \leq C_r\rho A_2 b^r. \tag{5.1}

For positive $r$, a further small factor is available, but the displayed bound suffices. The coefficient is independent of $\xi,u$. Since $R_2b\leq1$, (5.1) implies

c/M\in S(\rho A_2,g_{R_2}),\qquad c_z/M\in S(\rho A_2b,g_{R_2}). \tag{5.2}

The Poisson convention (1.3) gives

M^{-1}\{\xi c,f\} =(c/M)\partial_zf-(\xi c_z/M)\partial_\xi f. \tag{5.3}

For the first term, the retained derivative in (3.4), (5.2), and the product rule give

\begin{gathered} (c/M)\partial_zf \in S(\rho A_2B_2',g_{R_2}),\\ S(\rho A_2B_2',g_{R_2})\subset S(1,g_{R_2}), \end{gathered} \tag{5.4}

because $\rho A_2B_2'\leq\rho A_2B_2=1$.

For the second, the frequency factor $\xi/R_2$ and all its scaled derivatives are bounded on the cutoff support. The derivative $R_2\partial_\xi f$ belongs to $S(1,g_{R_2})$. Hence

\begin{gathered} (\xi c_z/M)\partial_\xi f \in S(\rho A_2b,g_{R_2}),\\ S(\rho A_2b,g_{R_2})\subset S(1,g_{R_2}), \end{gathered} \tag{5.5}

by $b\leq B_2$. Derivatives falling on $\xi$ are included: its first derivative is one and higher derivatives vanish, giving exactly the same scale bound. Derivatives falling on the coefficient use (5.1); no time derivative of the root scale is taken.

Combining (3.4), (4.4), (5.4)–(5.5) proves the primitive type-II cutoff and transport bounds in canonical coordinates. Equations (2.3)–(2.4) transfer them to the original coordinates.

6. Ordered products retain the bounds

Fix an index $j$. The cover theorem bounds the number of selected neighborhoods which meet $E_j$, not just the number through a single point. Consequently only a fixed number of the factors $1-\psi_i$, $i<j$, can be nonconstant on a neighborhood of $\operatorname{supp}\psi_j$. All other factors equal one there. The product-rule constants for the ordered component

\Psi_j=\psi_j\prod_{i<j}(1-\psi_i) \tag{6.1}

therefore depend on a fixed number of factors and the derivative order, independently of $j$. Their primitive $S(1,g_{R_2})$ bounds prove $\Psi_j\in S(1,g_{R_2})$.

The operator $\mathcal T$ is a derivation: both $D_t$ and the Hamiltonian vector field satisfy the product rule. Thus

\begin{aligned} M_j^{-1}\mathcal T\Psi_j ={}&(M_j^{-1}\mathcal T\psi_j) \prod_{i<j}(1-\psi_i)\\ &-\sum_{a<j}\frac{M_a}{M_j}\, \psi_j\,(M_a^{-1}\mathcal T\psi_a) \prod_{\substack{i<j\\ i\ne a}}(1-\psi_i). \end{aligned} \tag{6.2}

Each center scale is constant, so transverse derivatives do not act on $M_a/M_j$. Every nonzero term requires meeting supports, for which the cover theorem gives $M_a/M_j\leq C$. There are only a fixed number of such neighbors. Apply the finite product rule once more, using the primitive transport bounds. This proves $M_j^{-1}\mathcal T\Psi_j\in S(1,g_{R_2})$, with all constants uniform.

The sum $\Psi=\sum_j\Psi_j$ also belongs to $S(1,g_{R_2})$: at each point only a bounded number of component supports meet, so all its transverse derivatives are bounded by the corresponding finite sum. The component transport statement retains its own $M_j$, as in (1.4); no single global constant center scale is substituted for the family.

7. The small-gradient family uses its own radius

At a case-I center let $m$ be its scale. The original small-gradient derivative estimate holds on $|m(t-t_a)|<1,|w-w_a|<R$, and gives

|D_w^\beta q|\leq C_\beta\rho mR^{-|\beta|}. \tag{7.1}

The cutoff $\varphi$ is a fixed smooth function of $m(t-t_a)$ and $(w-w_a)/R_1$, supported where $|w-w_a|<R_1<R$. Thus

\varphi,\quad m^{-1}D_t\varphi\in S(1,g_{R_1}). \tag{7.2}

For a differentiated term of $m^{-1}\{q,\varphi\}$, split the additional derivatives between the differentiated symbol and the differentiated cutoff. Equation (7.1) gives

\begin{gathered} C_\alpha\rho R^{-1-|\beta|}R_1^{-1-|\alpha-\beta|}\\ =C_\alpha\frac{\rho}{RR_1} R_1^{-|\alpha|}(R_1/R)^{|\beta|}. \end{gathered} \tag{7.3}

The factor $R_1/R$ is smaller than one, and $\rho/(RR_1)\to0$ for fixed $\rho$. Summing the finite Poisson/product terms proves

\begin{gathered} m^{-1}H_q\varphi \in S\bigl(\rho/(RR_1),g_{R_1}\bigr),\\ S\bigl(\rho/(RR_1),g_{R_1}\bigr)\subset S(1,g_{R_1}). \end{gathered} \tag{7.4}

Together with (7.2), this proves the full type-I assertion and completes Theorem 1.1. ∎

All estimates concern actual cutoffs supported in the original symbol domain. The boundary-margin choice of type-II centers from the cover theorem supplies those supports without changing the localization region. The algebra uses the scalar derivative, coefficient and bracket bounds already proved in this course; it does not assume the later analytic estimates for localized operators.

8. Graded exercises with solutions

Exercise 1 — basic. Why does $g_r$ contain no $dt$ term? Does the uniformity in time in (1.2) nevertheless control $\partial_tf$?

Solution. This symbol class measures derivatives only in the transverse phase variables. Time is a parameter over which the constants are uniform. A uniformly bounded family can oscillate arbitrarily rapidly in time, so (1.2) alone gives no time-derivative bound. Here the explicit cutoff formula supplies $M^{-1}D_tf\in S(1,g_r)$ as a separate assertion.

Exercise 2 — intermediate. In (2.3), take $|\alpha|=4$ and $r=2$. Find the bound and explain what fails if $a\ll R_2$.

Solution. The term is bounded by $CA R_2^{-4}(R_2/a)^2$. With $a\geq R_2$, the last factor is at most one. With $a\ll R_2$, higher derivatives of the nonlinear map could amplify the symbol derivatives beyond the required $R_2$ scale. Uniform first-derivative bounds alone would not suffice.

Exercise 3 — intermediate. Verify both amplitudes in (5.4) and (5.5) when $\ell_0<R_2$, and when $\ell_0\geq R_2$.

Solution. In the first branch $B_2'=R_2^{-1}$, so $\rho A_2B_2'=\ell_0/R_2<1$. In the second $B_2'=B_2=\ell_0^{-1}$, so that amplitude is one. In either branch $b\leq B_2$, giving $\rho A_2b=\ell_0b\leq1$. The longer spatial width cancels the potentially large coefficient $\ell_0$ precisely where needed.

Exercise 4 — advanced. Derive (6.2) for two components, with $\Psi_2=\psi_2(1-\psi_1)$. Why is comparability of the center scales needed even if each primitive transport bound is known?

Solution. The derivation rule gives $\begin{aligned} M_2^{-1}\mathcal T\Psi_2 ={}&(M_2^{-1}\mathcal T\psi_2)(1-\psi_1)\\ &-\frac{M_1}{M_2}\psi_2 (M_1^{-1}\mathcal T\psi_1). \end{aligned} \tag{8.1}$ The primitive estimate for $\psi_1$ is normalized by $M_1$, while the component uses $M_2$. Their ratio must be bounded wherever both supports meet. The geometric overlap comparison supplies exactly that bound.

Exercise 5 — advanced. A family of supports has pointwise multiplicity at most two. Does that alone bound the number of supports which meet one fixed support? Give an example and identify the additional property proved for the adaptive cover.

Solution. One long interval can meet arbitrarily many pairwise disjoint short intervals; no point belongs to more than two intervals. Thus pointwise multiplicity alone does not bound the total number of neighbors. The adaptive cells have comparable center scales and tube widths whenever they meet, engulfing by a fixed enlargement, and disjoint small shrinks of comparable volume. Packing those shrinks gives the stronger bounded-neighbor property used for the finite product near a whole component support.

Exercise 6 — advanced. Let $r_1/r_2\to0$ and choose a fixed nonconstant $f\in C_c^\infty(\mathbb R)$. Show that $f(w/r_1)$ need not belong uniformly to $S(1,g_{r_2})$, although it belongs uniformly to $S(1,g_{r_1})$.

Solution. Choose $s$ where $f'(s)\ne0$. At $w=r_1s$, the derivative has magnitude $|f'(s)|/r_1$. A uniform $g_{r_2}$ bound would require this to be at most $C/r_2$, or $r_2/r_1\leq C/|f'(s)|$, which fails. Every derivative does satisfy its corresponding $r_1$ bound by rescaling. This explains why the two cutoff families retain distinct metrics.

References

The type-II and type-I cutoff-motion results are Lemmas 27.4.11 and 27.4.12 in Hörmander, The Analysis of Linear Partial Differential Operators IV, Chapter 27, Section 27.4. The proof above retains both metrics, all primitive and product transport terms, the residual small amplitude, the canonical composition estimates and the overlap scale ratios. The shared-flow covering argument needed for the product estimates is supplied by the preceding lesson.

Written by GPT-6.1 Sol (OpenAI), at Ultra reasoning effort, September 2026. Self-checked by the writing AI. Public domain (CC0).