Finding an admissible center nearby

A point in the large-gradient case need not be a good center for localization. Its bracket scale can fall sharply on a nearby cancellation plane. We will move by a bounded multiple of the smaller spatial radius to find a center whose entire prescribed neighborhood has a uniform lower bracket bound.

The proof minimizes the bracket scale on a closed ball. Two details matter: the minimum must stay in the large-gradient case, and a second cancellation plane must produce a point inside the original search ball. We verify both using the original transverse gradient and the uniform bounds for the canonical map.

Use the hypotheses and bracket scale from An adaptive scale for repeated brackets, the maps and estimates in An explicit canonical cell for a large transverse gradient, and the corrected transition in The linear frequency coefficient and the bracket scale. The geometric alternative and exact definition of an admissible neighborhood are proved in A cancellation plane or a uniform bracket lower bound.

1. The statement and the constants

Write \(w=(x',\xi')\), and set

The parameter \(\rho\geq1\) is fixed before \(\lambda\) is made small. Every constant in the conclusion is uniform in these parameters.

At a point \(p=(t,w)\), case II means that some low time-gradient violates the small-gradient threshold:

Case I is the complementary weak inequality for every such \(j\).

Theorem 1.1 (nearby admissible center). There is a uniform constant \(C_a\) such that every case-II point \((t_0,w_0)\) with

has an admissible center of type II at the same time, \((t_0,w_*)\), satisfying

The admissible neighborhood has the exact canonical-map image and the four coordinate restrictions in the preceding definition. Its center scale is \(\mu(t_0,w_*)\); comparability with \(\mu(t_0,w_0)\) is not part of the assertion.

We will use three uniform constants. Let \(C_\chi\geq1\) bound the first derivatives of the canonical maps and their inverses on the needed neighborhoods. Let \(C_\Xi\) give \(|\Xi|\leq C_\Xi R\) in the geometric alternative. Let \(K_0\) be the corrected case-transition constant: within the slightly smaller canonical cell, \(|\xi_2|\geq K_0R\) implies case I in the original coordinates. These constants are uniform over all centers under consideration. Choose

In particular \(C_a>C_\chi C_\Xi\).

2. The search ball and uniform local domains

The initial finite-bracket bound holds throughout the original domain and gives

The limits are uniform in the points in the original domain for each fixed \(\rho\). For example,

Let

Since \(\kappa<1\), \(\lambda R\to0\). The ball \(\mathcal K\) therefore lies inside \(|w|<3\lambda^{-1}/4\) for small \(\lambda\), by (1.3). Its points have a uniform margin inside the original symbol domain. The original time \(t_0\) also has the fixed margin in (1.3).

The local lemmas may consequently be applied with each point of \(\mathcal K\) as a translated center, with the same constants. Their constructions only need a fixed smaller original domain. The rescaled time interval has radius \(1/\mu\to0\). At a case-II center with selected gradient \(a\),

Thus every fixed multiple of the search radius is inside the canonical map's local domain, whose radius is a fixed multiple of \(a\). To check the image as well, substitute physical coordinates of size \(O(C_aR)\) into the explicit inverse formula from the canonical-cell construction. The distinguished spatial coordinate is recovered first; the remaining spatial and frequency coordinates are then recovered by bounded shears. All recovered coordinates have size \(O(C_aR)\ll a\), so both the map and inverse are defined there. Fixed linear canonical permutations made before constructing the map preserve the Euclidean norm and can be absorbed into \(C_\chi\).

The scale \(\mu(t_0,w)\) is continuous: it is the maximum of finitely many continuous absolute values followed by positive roots. It is positive by (2.1). Compactness of \(\mathcal K\) therefore gives a point \(w_*\in\mathcal K\) with

No smoothness of this maximum or of its roots is needed.

3. A nonadmissible starting center gives a smaller minimum

If \((t_0,w_0)\) is already an admissible center, take \(w_*=w_0\) and the theorem follows. Suppose it is not. Put \(M_0=\mu(t_0,w_0)\).

Apply the geometric alternative at the starting center, with local time origin \(t_0\) and canonical map \(\chi_0\) centered at \(w_0\). Since the lower-bound alternative does not hold there, the exact plane alternative gives, at local time zero and spatial coordinates zero,

The map notation includes the translation by \(w_0\) separately and \(\chi_0(0)=0\). Its Lipschitz bound gives

This point belongs to \(\mathcal K\). Hence the minimum in (2.5) satisfies

The geometric alternative used here is allowed to have overlapping conclusions. Nonadmissibility means that its lower-bound conclusion fails, so its plane conclusion still follows.

4. The minimizing center stays in case II

Choose \(s_0\leq\lfloor k/2\rfloor\) attaining the normalized gradient maximum at the original starting point. Put

The original second-transverse-derivative bound is uniform: \(|D_w^2\partial_t^{s_0}q|\leq C\). Integration along the segment inside \(\mathcal K\) gives

Since \(CC_aR/a_0\leq CC_aR^2/(\rho M_0)\to0\), make \(\lambda\) small enough that this ratio is less than \(1/4\). Then

On the other hand, (3.3) gives

Comparing (4.3) and (4.4) proves a strict violation of the case-I condition at \((t_0,w_*)\). Thus the minimizing point is a case-II center. This argument uses the original gradient and its Hessian; it does not assume that the root scale is differentiable or that the printed exact-radius transition is true.

5. A second cancellation plane contradicts minimality

Construct the exact canonical map \(\chi_*\) at \((t_0,w_*)\), using its selected gradient and scale \(m\). It fixes the local origin. Write

The inverse exists by (2.4). Its Lipschitz bound gives

With \(L_*=(\rho m)^{1/2}\) and \(b_*^{-1}\geq L_*\), equations (2.1) and (5.2) put this point in the slightly smaller canonical cell required by the corrected transition theorem, for small \(\lambda\). The original point \((t_0,w_0)\) is in case II. Therefore that theorem forces

This is the required frequency control. It uses a fixed sufficiently large multiple of \(R\), with no claim that the exact threshold is \(R\).

Suppose the minimizing center is not admissible. Its geometric alternative supplies a cancellation frequency \(\Xi_*\), with \(|\Xi_*|\leq C_\Xi R\), and a small bracket scale on the exact plane. Project the coordinates of the original starting point onto that plane:

The plane estimate applies at local time zero. Indeed the larger spatial radius in that estimate is at least \(R_2\); (5.2) and \(R/R_2\to0\) give \(|\bar z|<R_2\leq(B_{2,*}')^{-1}\) and \(|\bar u|<R_2\). Also \(|\Xi_*|=O(R)\ll L_*\), so the frequency belongs to the canonical-cell domain. Thus

The segment between \(\bar w\) and \(\widehat w\) stays in the canonical map's local domain, because both have size \(O(C_aR)\ll a_*\). Integrate its derivative along that segment. Equations (5.3)–(5.4) give

Therefore \(\widehat w_{\rm orig}\) is inside the original search ball \(\mathcal K\). This contradicts the defining minimum \(m\) in (2.5), since (5.5) is strictly smaller than \(m\). The minimizing center must be admissible. Its distance bound is already contained in \(w_*\in\mathcal K\), and its time coordinate remains \(t_0\). The theorem is proved. ∎

6. Graded exercises with solutions

Exercise 1 — basic. Explain why (1.3) uses half the original radius, rather than \(\lambda^{-1/2}\), and show that a search ball of any fixed radius \(C_aR\) fits inside the original domain for small \(\lambda\).

Solution. The inner radius is \(\tfrac12\lambda^{-1}\). If \(|w_0|<\tfrac12\lambda^{-1}\), then every point of the search ball has norm at most \(\tfrac12\lambda^{-1}+C_a\lambda^{-\kappa}\). Multiplying by \(\lambda\) gives \(\tfrac12+C_a\lambda^{1-\kappa}<\tfrac34\) for small \(\lambda\), because \(\kappa<1\). The smaller radius \(\lambda^{-1/2}\) is a different restriction and is not needed.

Exercise 2 — intermediate. Suppose a selected gradient has norm \(a_0\), its change across the ball is below \(a_0/4\), and the new scale is below \(M_0/2\). Verify the strict case-II conclusion for every possible selected index \(s_0\geq0\).

Solution. The new gradient norm is greater than \(3a_0/4\). Its case-I threshold is smaller than \(a_0/2^{s_0+1}\leq a_0/2\), by the original strict case-II inequality. Since \(3/4>1/2\), it violates that threshold for all selected indices, including \(s_0=0\).

Exercise 3 — intermediate. A canonical inverse sends the original point to \((\bar z,\bar\xi_2,\bar u)\), with \(|\bar\xi_2|<K_0R\). The cancellation frequency satisfies \(|\Xi_*|\leq C_\Xi R\), and the canonical map has Lipschitz constant \(C_\chi\). Compute an upper bound for the physical displacement after projecting to \(\xi_2=-\Xi_*\).

Solution. The canonical displacement has only one nonzero coordinate and magnitude \(|\bar\xi_2+\Xi_*|\leq(K_0+C_\Xi)R\). The physical displacement is at most \(C_\chi(K_0+C_\Xi)R\). Choosing \(C_a\) larger than that coefficient places the projected point inside the original search ball, regardless of whether the minimizing center itself lies on the ball's boundary.

Exercise 4 — advanced. Fix \(\rho\geq1\), set \(a=\lambda^{-1}\), and take \(k=1\), \(q(t,z,\xi)=a^2t+a\xi+3aR/4\). For sufficiently small \(\lambda\), find the center scale at \((t,z,\xi)=(0,0,0)\), and the scale at \((0,0,-3R/4)\). Determine the gradient cases and explain why the theorem does not compare the two scales by a fixed constant.

Solution. The value at the original center dominates the time-bracket root once \(R\) is large. Thus

Both centers are in case II. At the second point the symbol value is zero and the constant time bracket \(a^2\) gives the displayed scale. At every point of any neighborhood the bracket scale is at least \(m\), so this second center satisfies the uniform lower alternative and is admissible. The displacement is \(3R/4\). Yet \(m/M_0=4\sqrt\rho/(3R)\to0\), showing that proximity alone does not give comparable center scales. The symbol has the required derivative bounds, global finite-bracket lower bound and positive time slope.

Exercise 5 — advanced. Why is it necessary to keep the corrected constant \(K_0\) in the choice of \(C_a\)? Could the proof simply replace it by one to make the search radius smaller?

Solution. The original case-II point need only satisfy \(|\bar\xi_2|<K_0R\) in the minimizing center's coordinates. Entry into case I at the exact radius \(R\) is false under the hypotheses, as the preceding coefficient lesson's admissible countermodel shows. The triangle inequality in (5.6) must retain \(K_0\). Its value is a fixed uniform constant, so including it changes neither the \(O(R)\) conclusion nor the order of parameter choices.

Exercise 6 — advanced. Is it legitimate to differentiate the minimizing root scale to obtain an Euler equation at \(w_*\)? Give the exact properties of the scale that the proof actually uses.

Solution. The scale is a finite maximum of absolute values raised to positive roots. A bracket can vanish and different words can attain the maximum, so differentiability is not guaranteed. Moreover the minimum can occur on the boundary of the search ball. The proof uses continuity to attain a positive minimum, the explicit lower finite-type bound, the original smooth gradient/Hessian estimates, canonical invariance and strict inequalities from the cancellation-plane alternative. No Euler equation or derivative of the root scale is needed.

References

The nearby-center theorem follows the local analysis in Hörmander, The Analysis of Linear Partial Differential Operators IV, Chapter 27, Section 27.4, especially Proposition 27.4.10. The proof above retains the full inner domain and proves case-II persistence and search-ball containment explicitly. It uses the corrected large-frequency transition from the preceding coefficient lesson, with its uniform constant absorbed into \(C_a\). Covering, overlap and cutoff-motion estimates are subsequent results.

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