# AMRAL × Lebesgue Universal Covering — Round 02
## Curvature-Density Finite Compiler and Certified Shape Approximation

**Document code:** AMRAL-LUC-FC-R02  
**Version:** v0.1  
**Date:** 2026-09-18  
**Research status:** Round 02 / Formal push / Constructive finite-shape certificate  
**Research mode:** Human-Directed + Semi-Autonomous AI Mathematical Research  
**Research initiation and methodology source:** Neo.K  
**AI collaborating researcher and primary executor:** Aletheia / ChatGPT, GPT-5.6 Sol  
**Parent methodology:** Relational Constraint–Handoff Methodology (RCHM)  
**Prerequisite documents:** AMRAL-LUC-FC-R00 v0.2; AMRAL-LUC-FC-R01 v0.1  

---

# 0. This Round's Summary Verdict

Round 01 strictly compressed the original target family down to centered unit constant-width bodies, and obtained the canonical curvature-density domain:

$$
\mathcal R
=
\left\{
r\in L^\infty(S^1):
\begin{array}{l}
0\le r\le1,\\
r(\theta+\pi)=1-r(\theta),\\
\int r\cos=0,\\
\int r\sin=0
\end{array}
\right\}.
$$

Round 01's next exact gate was:

> Can this still infinite-dimensional legal domain be constructed into a **finite, verifiable dictionary in which every element preserves convexity and constant width**, together with an explicit Hausdorff error?

This round completes the answer:

$$
\boxed{\text{YES}.}
$$

The core results are:

1. For the support function $h$ of any centered unit constant-width body, taking a circular convolution with a squared-Fejér / Jackson-type positive kernel yields a finite-degree trigonometric polynomial:

$$
\mathcal J_Nh,
$$

which **exactly preserves**:
- convexity;
- constant width $1$;
- the Steiner gauge;
- reflection legality;
- the curvature-density box constraint.

2. This round independently derives an explicit Hausdorff error:

$$
\boxed{
\|\mathcal J_Nh-h\|_\infty
\le
E_N
:=
\frac{
\pi^2(4N+1)
}{
2(2N^2+4N+3)(N+1)
}.
}
$$

Hence:

$$
E_N
=
O(N^{-2}).
$$

3. Simple Fourier-coefficient rounding can still break convexity by itself, so this round adds a safety interiorization:

$$
h_{N,\delta}
=
(1-\delta)\mathcal J_Nh
+
\frac{\delta}{2}.
$$

Its curvature density stays away from the boundary values $0,1$ by a safety distance of at least:

$$
\frac{\delta}{2}.
$$

4. For the odd Fourier coefficients of degree $\le2N$, lattice quantization is applied; if the grid spacing $q$ satisfies:

$$
qA_N
\le
\frac{\delta}{2},
$$

where:

$$
A_N
=
\frac43(N-1)N(N+1),
$$

then the quantized result is still a legal unit constant-width body.

5. Therefore a **genuinely finite legal dictionary** can be established:

$$
\mathscr D_{N,\delta,q},
$$

such that for every centered unit constant-width body $K$ there exists:

$$
\widehat K
\in
\mathscr D_{N,\delta,q}
$$

satisfying:

$$
\boxed{
d_H(K,\widehat K)
\le
E_N
+
\frac{\delta}{2}
+
(N-1)q.
}
$$

6. Combined with Round 01's shape-Lipschitz theorem, the finite dictionary's error can be transferred directly to the placement functional.

Hence Round 02's core gate verdict is:

$$
\boxed{
\text{CONSTRUCTIVE FINITE SHAPE COMPILER: CLOSED}
}
$$

but:

$$
\boxed{
\text{GLOBAL FINITE CLOSURE: STILL OPEN}
}
$$

because the remaining continuous degrees of freedom still include:

- orientation;
- support direction;
- the candidate-cover boundary;
- active-branch saturation.

---

# 1. A Necessary Strengthening of Round 01: Non-Smooth Constant-Width Bodies Also Legally Enter the $r$ Domain

Round 01 used:

$$
r=h+h''
$$

and wrote $r$ as:

$$
0\le r\le1.
$$

This is intuitive for smooth curves, but the Lebesgue problem cannot be allowed to quietly exclude non-smooth constant-width bodies such as the Reuleaux triangle.

So this round first upgrades this step to the distribution / measure level.

---

# 2. The Curvature-Measure Version

Let $h$ be the $2\pi$-periodic support function of an arbitrary planar compact convex body.

In the distributional sense:

$$
\mu_h
=
h+h''
$$

is a non-negative finite Borel measure.

For a unit constant-width body:

$$
h(\theta)+h(\theta+\pi)=1.
$$

Applying the operator:

$$
D^2+1
$$

gives:

$$
\mu_h
+
\tau_\pi\mu_h
=
d\theta,
$$

where $\tau_\pi$ denotes rotation by angle $\pi$.

Because:

$$
\mu_h\ge0,
\qquad
\tau_\pi\mu_h\ge0,
$$

and the two sum to the Lebesgue measure, it follows that:

$$
\mu_h
\ll
d\theta.
$$

Hence there exists:

$$
r\in L^\infty(S^1)
$$

such that:

$$
d\mu_h
=
r(\theta)d\theta.
$$

And since:

$$
\mu_h
\le
d\theta,
$$

we obtain:

$$
0\le r(\theta)\le1
$$

almost everywhere.

At the same time:

$$
r(\theta+\pi)
=
1-r(\theta)
$$

almost everywhere.

So Round 01's curvature-density representation applies not only to $C^2$ bodies, but also covers Reuleaux-type non-smooth constant-width bodies.

Status:

`R-DOMAIN NONSMOOTH AUDIT: PASS`

---

# 3. Circular Convolution Convention

For $2\pi$-periodic functions or measures, define the normalized circular convolution:

$$
(K*f)(\theta)
=
\frac1{2\pi}
\int_{-\pi}^{\pi}
K(t)f(\theta-t)\,dt.
$$

If:

$$
K(t)\ge0
$$

and:

$$
\frac1{2\pi}
\int_{-\pi}^{\pi}
K(t)\,dt
=
1,
$$

then $K*$ is a convex averaging operator.

---

# 4. Positive-Kernel Legality Theorem

## Theorem 4.1

Let $h$ be the support function of a centered unit constant-width body.

Let $K$ be a non-negative, normalized, even $2\pi$-periodic kernel.

Define:

$$
h_K
=
K*h.
$$

Then $h_K$ is still the support function of a centered unit constant-width body.

### Proof: constant width

From:

$$
h(\theta)+h(\theta+\pi)=1,
$$

we have:

$$
\begin{aligned}
h_K(\theta)+h_K(\theta+\pi)
&=
K*
\left(
h(\theta)+h(\theta+\pi)
\right)\\
&=
K*1\\
&=
1.
\end{aligned}
$$

---

### Proof: convexity

In the distributional sense:

$$
h+h''
=
r
\ge0.
$$

Convolution commutes with differentiation, so:

$$
h_K+h_K''
=
K*(h+h'')
=
K*r.
$$

Because:

$$
K\ge0,
\qquad
r\ge0,
$$

we have:

$$
K*r\ge0.
$$

Hence $h_K$ is the support function of some convex body.

---

### Proof: curvature upper bound

From:

$$
0\le r\le1
$$

and positive normalized averaging:

$$
0
\le
K*r
\le
1.
$$

---

### Proof: Steiner gauge

For a centered body, the first harmonic is zero.

In Fourier space, convolution only multiplies by a scalar multiplier and does not create a frequency that was not already present, so the first harmonic remains zero.

Hence:

$$
s(K)=0.
$$

Q.E.D.

---

# 5. The Fejér Kernel

Define the standard Fejér kernel:

$$
F_N(t)
=
\frac1{N+1}
\left(
\frac{
\sin\left(\frac{N+1}{2}t\right)
}{
\sin\left(\frac t2\right)
}
\right)^2.
$$

Take the continuous extension at $t=0$:

$$
F_N(0)=N+1.
$$

It satisfies:

$$
F_N(t)\ge0,
$$

$$
F_N(-t)=F_N(t),
$$

$$
\frac1{2\pi}
\int_{-\pi}^{\pi}
F_N(t)\,dt
=
1.
$$

Its Fourier expansion is:

$$
F_N(t)
=
\sum_{|k|\le N}
\left(
1-\frac{|k|}{N+1}
\right)
e^{ikt}.
$$

---

# 6. The Squared-Fejér / Jackson-Type Kernel

Let:

$$
S_N
=
\frac1{2\pi}
\int_{-\pi}^{\pi}
F_N(t)^2\,dt.
$$

By Parseval:

$$
S_N
=
1
+
2
\sum_{k=1}^{N}
\left(
1-\frac{k}{N+1}
\right)^2.
$$

Computing this gives:

$$
\boxed{
S_N
=
\frac{
2N^2+4N+3
}{
3(N+1)
}.
}
$$

Define:

$$
\boxed{
J_N(t)
=
\frac{
F_N(t)^2
}{
S_N
}.
}
$$

Then:

$$
J_N\ge0,
$$

$$
J_N(-t)=J_N(t),
$$

$$
\frac1{2\pi}
\int J_N
=
1.
$$

Since $F_N$ has degree $N$:

$$
J_N
$$

is a non-negative trigonometric polynomial of degree:

$$
2N.
$$

Define the first layer of the Round 02 compiler:

$$
\boxed{
\mathcal J_Nh
=
J_N*h.
}
$$

By Theorem 4.1:

$$
\mathcal J_Nh
$$

is always a legal centered unit constant-width support function.

---

# 7. Why We Can Obtain $O(N^{-2})$

For a centered unit constant-width body:

$$
0\in K
$$

because the Steiner point lies inside the convex body.

Also:

$$
\operatorname{diam}(K)=1.
$$

So:

$$
K\subseteq\overline B(0,1).
$$

Hence:

$$
0\le h(\theta)\le1.
$$

On the other hand:

$$
h''=r-h
$$

almost everywhere.

From:

$$
0\le r\le1,
\qquad
0\le h\le1,
$$

we obtain:

$$
\boxed{
\|h''\|_\infty\le1.
}
$$

So $h$ belongs to:

$$
W^{2,\infty}(S^1).
$$

This is stronger regularity than mere Lipschitz continuity.

---

# 8. The Even-Kernel Second-Difference Inequality

For any:

$$
h\in W^{2,\infty},
$$

we have:

$$
|h(\theta+t)+h(\theta-t)-2h(\theta)|
\le
\|h''\|_\infty t^2.
$$

Since $J_N$ is even:

$$
\begin{aligned}
\mathcal J_Nh(\theta)-h(\theta)
&=
\frac1{4\pi}
\int_{-\pi}^{\pi}
J_N(t)
[
h(\theta+t)+h(\theta-t)-2h(\theta)
]
\,dt.
\end{aligned}
$$

Hence:

$$
\boxed{
\|\mathcal J_Nh-h\|_\infty
\le
\frac{\|h''\|_\infty}{2}
\mu_{2,N},
}
$$

where:

$$
\mu_{2,N}
=
\frac1{2\pi}
\int_{-\pi}^{\pi}
t^2J_N(t)\,dt.
$$

---

# 9. An Explicit Upper Bound for the Second Moment of $J_N$

For:

$$
0<|t|\le\pi,
$$

we have:

$$
F_N(t)\le N+1
$$

as well as:

$$
F_N(t)
\le
\frac{
\pi^2
}{
(N+1)t^2
}.
$$

Let:

$$
t_0
=
\frac{\pi}{N+1}.
$$

Then:

$$
\begin{aligned}
\mu_{2,N}
&=
\frac1{\pi S_N}
\int_0^\pi
t^2F_N(t)^2\,dt\\
&\le
\frac1{\pi S_N}
\left[
\int_0^{t_0}
t^2(N+1)^2\,dt
+
\int_{t_0}^{\pi}
t^2
\frac{\pi^4}{(N+1)^2t^4}
\,dt
\right].
\end{aligned}
$$

The first term:

$$
\int_0^{t_0}
t^2(N+1)^2\,dt
=
\frac{\pi^3}{3(N+1)}.
$$

The second term:

$$
\int_{t_0}^{\pi}
\frac{\pi^4}{(N+1)^2t^2}
\,dt
=
\frac{\pi^3N}{(N+1)^2}.
$$

Substituting:

$$
S_N
=
\frac{2N^2+4N+3}{3(N+1)},
$$

gives:

$$
\boxed{
\mu_{2,N}
\le
\frac{
\pi^2(4N+1)
}{
(2N^2+4N+3)(N+1)
}.
}
$$

And since:

$$
\|h''\|_\infty\le1,
$$

we get:

$$
\boxed{
\|\mathcal J_Nh-h\|_\infty
\le
E_N
:=
\frac{
\pi^2(4N+1)
}{
2(2N^2+4N+3)(N+1)
}.
}
$$

and:

$$
E_N
=
O(N^{-2}).
$$

This is Round 02's first constructive quantitative theorem.

---

# 10. Relationship to the External 2026 Finite Hierarchy

The Reuleaux-type hierarchy proposed by Zeng in 2026 gives:

$$
0
\le
a_{\mathrm{Leb}}-\Lambda_M
\le
CM^{-2}.
$$

This round's:

$$
E_N=O(N^{-2})
$$

is **not the same theorem** as that one.

The difference:

- What Zeng controls is the finite-arc variational hierarchy of the Lebesgue universal-cover constant;
- What Round 02 controls is the Hausdorff approximation error, after passing through the legal positive-kernel compiler, of an arbitrary constant-width target shape.

So this round cannot claim to reproduce or replace the Zeng hierarchy.

Its real new use is:

> We obtain a Fourier / positive-kernel route that is different from the Reuleaux finite-arc route, and it can serve as a second finite-ization branch and cross-validator for AMRAL.

---

# 11. The First-Layer Compiler Is Not Yet a True Finite Dictionary

Although:

$$
\mathcal J_Nh
$$

is a finite-dimensional trigonometric polynomial of degree $\le2N$, its coefficients still take continuum values.

So:

$$
\boxed{
\text{finite-dimensional}
\neq
\text{finite dictionary}.
}
$$

RCHM does not permit claiming finite closure prematurely at this step.

So a second layer is still needed.

---

# 12. Safety Interiorization

Let:

$$
0<\delta<1.
$$

Define:

$$
\boxed{
h_{N,\delta}
=
(1-\delta)\mathcal J_Nh
+
\frac{\delta}{2}.
}
$$

Geometrically this equals the Minkowski convex combination:

$$
K_{N,\delta}
=
(1-\delta)K_N
+
\delta B_{1/2}.
$$

Since a disc of radius $1/2$ is a unit constant-width body:

$$
h_{N,\delta}
$$

is still a centered unit constant-width support function.

Its curvature density:

$$
r_{N,\delta}
=
(1-\delta)\mathcal J_Nr
+
\frac{\delta}{2}.
$$

Because:

$$
0\le\mathcal J_Nr\le1,
$$

we obtain:

$$
\boxed{
\frac{\delta}{2}
\le
r_{N,\delta}
\le
1-\frac{\delta}{2}.
}
$$

This:

$$
\frac{\delta}{2}
$$

is exactly the safety margin for coefficient quantization.

---

# 13. Interiorization Error

Because:

$$
0\le\mathcal J_Nh\le1,
$$

we have:

$$
\left|
\frac12
-
\mathcal J_Nh
\right|
\le
\frac12.
$$

Hence:

$$
\boxed{
\|h_{N,\delta}-\mathcal J_Nh\|_\infty
\le
\frac{\delta}{2}.
}
$$

Combining:

$$
\boxed{
\|h_{N,\delta}-h\|_\infty
\le
E_N+\frac{\delta}{2}.
}
$$

---

# 14. The Odd Finite Fourier Form

Because the original centered constant-width support function has only:

- a constant mode $1/2$;
- odd modes;
- no first harmonic;

and convolution does not add new frequencies:

$$
h_{N,\delta}(\theta)
=
\frac12
+
\sum_{j=1}^{N-1}
\left[
a_j\cos((2j+1)\theta)
+
b_j\sin((2j+1)\theta)
\right].
$$

The highest frequency:

$$
2N-1.
$$

So only:

$$
2(N-1)
$$

real coefficients remain.

---

# 15. Coefficient Quantization

Choose a grid spacing:

$$
q>0.
$$

For each coefficient take the nearest lattice point:

$$
\widehat a_j
\in
q\mathbb Z,
$$

$$
\widehat b_j
\in
q\mathbb Z,
$$

such that:

$$
|\widehat a_j-a_j|
\le
\frac q2,
$$

$$
|\widehat b_j-b_j|
\le
\frac q2.
$$

Define:

$$
\widehat h
=
\frac12
+
\sum_{j=1}^{N-1}
\left[
\widehat a_j\cos((2j+1)\theta)
+
\widehat b_j\sin((2j+1)\theta)
\right].
$$

Since it still has only odd modes and no first harmonic:

$$
\widehat h(\theta)
+
\widehat h(\theta+\pi)
=
1.
$$

The Steiner gauge is also preserved exactly.

---

# 16. The Convexity Error of Quantization

Let:

$$
k_j=2j+1.
$$

Then the curvature perturbation:

$$
\Delta r
=
(\widehat h-h_{N,\delta})
+
(\widehat h-h_{N,\delta})''.
$$

So:

$$
\begin{aligned}
\|\Delta r\|_\infty
&\le
\sum_{j=1}^{N-1}
(k_j^2-1)
\left(
|\Delta a_j|
+
|\Delta b_j|
\right)\\
&\le
q
\sum_{j=1}^{N-1}
(k_j^2-1).
\end{aligned}
$$

And:

$$
k_j^2-1
=
(2j+1)^2-1
=
4j(j+1).
$$

Hence:

$$
A_N
:=
\sum_{j=1}^{N-1}
(k_j^2-1)
$$

satisfies:

$$
\boxed{
A_N
=
\frac43
(N-1)N(N+1).
}
$$

So:

$$
\boxed{
\|\Delta r\|_\infty
\le
qA_N.
}
$$

---

# 17. The Legality-Preserving Quantization Theorem

If:

$$
\boxed{
qA_N
\le
\frac{\delta}{2},
}
$$

then:

$$
0
\le
\widehat r
=
\widehat h+\widehat h''
\le
1.
$$

Hence:

$$
\boxed{
\widehat h
}
$$

is still the support function of a legal centered unit constant-width body.

This step closes the warning Round 01 raised:

> naive Fourier truncation / rounding can break convexity.

This round does not assume positivity, but rather guarantees positivity by means of an interior margin plus an explicit coefficient bound.

---

# 18. The Support Error of Quantization

We have:

$$
\begin{aligned}
\|\widehat h-h_{N,\delta}\|_\infty
&\le
\sum_{j=1}^{N-1}
\left(
|\Delta a_j|
+
|\Delta b_j|
\right)\\
&\le
(N-1)q.
\end{aligned}
$$

Hence the total error:

$$
\boxed{
\|\widehat h-h\|_\infty
\le
E_N
+
\frac{\delta}{2}
+
(N-1)q.
}
$$

For compact convex bodies:

$$
d_H(K,L)
=
\|h_K-h_L\|_\infty.
$$

So:

$$
\boxed{
d_H(K,\widehat K)
\le
E_N
+
\frac{\delta}{2}
+
(N-1)q.
}
$$

---

# 19. A Genuinely Finite Legal Dictionary

For centered unit constant-width support functions:

$$
0\le h\le1.
$$

Hence each Fourier coefficient has a fixed uniform bound; for instance one may safely take:

$$
|a_j|,
|b_j|
<
2.
$$

If:

$$
q\le\frac12,
$$

the nearest lattice point can still be confined to:

$$
[-2,2].
$$

Define:

$$
\mathscr D_{N,\delta,q}
$$

as the set of all forms:

$$
\widehat h(\theta)
=
\frac12
+
\sum_{j=1}^{N-1}
[
\widehat a_j\cos(k_j\theta)
+
\widehat b_j\sin(k_j\theta)
]
$$

where:

$$
\widehat a_j,\widehat b_j
\in
q\mathbb Z\cap[-2,2],
$$

and:

$$
0
\le
\widehat h+\widehat h''
\le
1.
$$

Then:

$$
\mathscr D_{N,\delta,q}
$$

is a finite set.

Its rough cardinality bound:

$$
\boxed{
|\mathscr D_{N,\delta,q}|
\le
\left(
\left\lceil\frac4q\right\rceil+1
\right)^{2(N-1)}.
}
$$

This bound is extremely large, but it proves that:

$$
\boxed{
\text{a finite dictionary genuinely exists, and can be enumerated.}
}
$$

What is being pursued here is proof-level finiteness, not computational efficiency.

---

# 20. The Constructive Finite Legal Dictionary Theorem

## Theorem 20.1

For any:

$$
N\ge2,
$$

$$
0<\delta<1,
$$

and:

$$
0<q
\le
\min
\left\{
\frac12,
\frac{\delta}{2A_N}
\right\},
$$

every centered unit constant-width body $K$ has some:

$$
\widehat K
\in
\mathscr D_{N,\delta,q}
$$

such that:

$$
\boxed{
d_H(K,\widehat K)
\le
\varepsilon_{N,\delta,q},
}
$$

where:

$$
\boxed{
\varepsilon_{N,\delta,q}
=
E_N
+
\frac{\delta}{2}
+
(N-1)q.
}
$$

And:

$$
\varepsilon_{N,\delta,q}
\to0
$$

can be achieved through:

$$
N\to\infty,
\qquad
\delta\to0,
\qquad
q\to0
$$

with a suitable coupling.

Q.E.D.

---

# 21. An Explicit Choice of Parameters for Arbitrary $\varepsilon$

Given:

$$
0<\varepsilon<1.
$$

First choose $N$ such that:

$$
E_N
\le
\frac{\varepsilon}{3}.
$$

Take:

$$
\delta
=
\frac{\varepsilon}{3}.
$$

Then choose:

$$
q
\le
\min
\left\{
\frac{\delta}{2A_N},
\frac{\varepsilon}{3(N-1)}
\right\}.
$$

Then:

$$
\frac{\delta}{2}
=
\frac{\varepsilon}{6},
$$

and:

$$
(N-1)q
\le
\frac{\varepsilon}{3}.
$$

So:

$$
d_H(K,\widehat K)
\le
\frac56\varepsilon
<
\varepsilon.
$$

Hence:

$$
\boxed{
\forall\varepsilon>0,
\quad
\exists
\text{ a finite legal dictionary }
\mathscr D_\varepsilon
}
$$

that is a Hausdorff $\varepsilon$-net for the whole family of centered unit constant-width bodies.

This formally upgrades Round 01's:

`ABSTRACT FINITE NET`

to:

`CONSTRUCTIVE LEGAL FINITE NET`

---

# 22. A Numerical-Scale Example

This round's explicit kernel bound:

$$
E_N
=
\frac{
\pi^2(4N+1)
}{
2(2N^2+4N+3)(N+1)
}
$$

gives:

| $N$ | Upper bound on $E_N$ |
|---:|---:|
| $8$ | $0.1110078205$ |
| $16$ | $0.0325878434$ |
| $32$ | $0.0088529557$ |
| $64$ | $0.0023087740$ |
| $128$ | $0.0005896237$ |
| $256$ | $0.0001489913$ |

These are proof-level coarse bounds and do not represent how large the actual smoothing error will be in practice.

In this round's sanity example, the actual error at $N=8$ is approximately:

$$
0.00639,
$$

markedly smaller than the theoretical upper bound:

$$
0.11101.
$$

So there remains substantial room for optimizing the constants in the future.

---

# 23. Signed Placement Margin

Round 01 used the non-negative directed-distance functional:

$$
\Psi_U(K)\ge0.
$$

To let the finite net perform universal-cover certification directly, this round introduces a signed margin:

$$
\boxed{
M_U(K)
=
\min_{Q\in O(2),\,t\in\mathbb R^2}
\max_{u\in S^1}
[
h_K(Q^Tu)
+t\cdot u
-h_U(u)
].
}
$$

Then:

$$
M_U(K)\le0
$$

if and only if:

$$
K
$$

can be placed inside $U$ by a congruence.

If:

$$
M_U(K)<0,
$$

this indicates that uniform support slack exists.

---

# 24. The Shape-Lipschitz Property of the Signed Margin

For any centered constant-width $K,L$:

$$
\boxed{
|M_U(K)-M_U(L)|
\le
d_H(K,L).
}
$$

The proof has exactly the same form as Round 01's proof for $\Psi_U$:

For any fixed $(Q,t)$:

$$
\left|
\max_u R_K(u)
-
\max_u R_L(u)
\right|
\le
\|h_K-h_L\|_\infty.
$$

Then taking the minimum over configurations gives the result.

---

# 25. The Finite Dictionary Certification Sandwich

Let:

$$
\mathscr D
=
\mathscr D_{N,\delta,q}
$$

be an $\varepsilon$-net, where:

$$
\varepsilon
=
\varepsilon_{N,\delta,q}.
$$

Then:

$$
\boxed{
\max_{\widehat K\in\mathscr D}
M_U(\widehat K)
\le
\max_{K\in\mathcal W_1^0}
M_U(K)
\le
\max_{\widehat K\in\mathscr D}
M_U(\widehat K)
+\varepsilon.
}
$$

Hence there are two direct certificate rules.

---

## Rule A: Finite Universal Certificate with Slack

If:

$$
\boxed{
\max_{\widehat K\in\mathscr D}
M_U(\widehat K)
\le
-\varepsilon,
}
$$

then:

$$
\max_{K\in\mathcal W_1^0}
M_U(K)
\le0.
$$

So:

$$
\boxed{
U\text{ is a universal cover}.
}
$$

---

## Rule B: Finite Non-Universality Witness

If there exists:

$$
\widehat K\in\mathscr D
$$

such that:

$$
M_U(\widehat K)>0,
$$

then $\widehat K$ is itself a legal constant-width witness.

So:

$$
\boxed{
U\text{ is not universal}.
}
$$

---

## The Ambiguous Strip

If:

$$
-\varepsilon
<
\max_{\mathscr D}M_U
\le0,
$$

the finite dictionary alone is not enough to decide.

At this point one must:

- increase $N$;
- decrease $\delta$;
- decrease $q$;
- or apply local refinement to the active branch.

This is exactly the entry point for a future adaptive certificate.

---

# 26. What This Round Has Actually Closed

After Round 01 there still remained an escape:

$$
r\in\mathcal R
$$

's infinite-dimensional microstructure.

Round 02 now proves that for any accuracy:

$$
\varepsilon>0,
$$

this microstructure can be controlled by a **finite legal dictionary**, and every element of the dictionary is not a heuristic shape but a genuine unit constant-width convex body.

Hence:

$$
\boxed{
\text{target-shape infinity}
}
$$

has been downgraded from:

`UNCONTROLLED INFINITE DOMAIN`

to:

`CERTIFIED FINITE-RESOLUTION DOMAIN`

---

# 27. But This Is Still Not Global Finite Closure

RCHM must continue to guard against overclaiming.

Even though:

$$
\mathscr D_\varepsilon
$$

is finite, for every dictionary shape we still face the continuous placement problem:

$$
\min_{Q,t}
\max_{\theta}
$$

So the remaining degrees of freedom include:

$$
Q\in O(2),
$$

$$
t\in\mathbb R^2,
$$

$$
\theta\in S^1.
$$

Round 01 already compressed fixed-$Q$ translation down to at most three active directions, but:

- the orientation continuum has not yet been finite-certified;
- the support-direction continuum has not yet been finite-certified;
- active-triple switching has not yet been saturated.

Hence:

$$
\boxed{
\text{finite shape dictionary}
\not\Rightarrow
\text{finite global proof}.
}
$$

---

# 28. Relationship to the Reuleaux-Only Route

Mishra 2026's new lower bound uses:

- the disc;
- the Reuleaux triangle;
- the Reuleaux pentagon;

and constructs a massive finite certificate over the placement space.

Zeng 2026 instead establishes a Reuleaux-type finite-arc hierarchy.

This round's compiler plays a different role:

$$
\boxed{
\text{all constant-width bodies}
\to
\text{a finite legal Fourier dictionary}.
}
$$

This provides a second representation that can run in parallel with the Reuleaux hierarchy.

In the future, if the two finite routes both converge to the same active structure, confidence will increase substantially.

If they differ, the difference itself will expose:

- branch omission;
- representation bias;
- saturation failure.

This is very well suited to AMRAL's dual-route / multi-AI verification.

---

# 29. The RCHM Freedom Ledger

Round 01's residual:

$$
\mathcal F_1
=
\{
\text{shape microstructure},
\text{high odd modes},
\text{rotation},
\text{translation},
\text{support direction},
\text{active switching},
\text{container boundary}
\}.
$$

For the target-shape part, Round 02 completes:

$$
\text{shape microstructure}
\to
\text{a finite legal codebook at resolution }\varepsilon.
$$

as well as:

$$
\text{high odd modes}
\to
\text{explicit }E_N\text{ tail control}.
$$

So:

$$
\mathcal F_2
=
\{
\text{rotation},
\text{translation},
\text{support direction},
\text{active switching},
\text{container boundary},
\text{resolution limit }\varepsilon\to0
\}.
$$

where translation, under fixed orientation, has already been locally finite-actived.

---

# 30. The Escape Ledger

## E1: Orientation Continuum

There is still:

$$
\phi\in S^1.
$$

## E2: Support-Direction Continuum

Every placement still has to handle:

$$
\max_{\theta\in S^1}.
$$

## E3: Active-Set Bifurcation

Translation's optimum requires at most three active directions, but these three directions can switch with:

$$
(\widehat K,\phi,U).
$$

## E4: Zero-Slack Limit

For a genuine extremal universal cover, the active shapes very likely satisfy:

$$
M_U(K)=0.
$$

So a single coarse finite dictionary's negative-slack rule is not enough to complete the final exact proof.

## E5: Candidate-Cover Infinity

$U$ itself is still an infinite-dimensional convex body.

---

# 31. COMPUTE-DEFERRED

This round's core theorems do not depend on large-scale computation.

But the following tasks have now been made concrete.

## C02-1: Dictionary Growth Census

Input:

$$
(N,\delta,q).
$$

Output:

- the number of legal coefficient cells;
- the pruning ratio;
- the curvature-positivity pass rate;
- the symmetry classes.

## C02-2: Jackson-Compiler Adversarial Benchmark

Compare:

- the Reuleaux triangle;
- the Reuleaux pentagon;
- high-mode constant-width bodies;
- random legal $r$;

at different $N$, comparing:

$$
d_H(K,\mathcal J_NK)
$$

against the theoretical:

$$
E_N.
$$

## C02-3: Coefficient-Quantization Branch Compression

Study whether the legal coefficient grid can be greatly shrunk using:

- interval pruning;
- symmetry quotients;
- active-direction signatures.

All of these are marked:

`COMPUTE-DEFERRED`

---

# 32. This Round's Sanity Validation

Using the test family:

$$
r(\theta)
=
\frac12
+
0.25\cos3\theta
+
0.15\sin5\theta.
$$

Its range:

$$
0.1076858\ldots
\le
r
\le
0.8923141\ldots
$$

is legal.

Reconstruction:

$$
h(\theta)
=
\frac12
+
\frac{0.25}{1-9}\cos3\theta
+
\frac{0.15}{1-25}\sin5\theta.
$$

Take:

$$
N=8,
\qquad
\delta=0.1.
$$

The experiment obtains:

- positive-kernel normalization: PASS;
- actual smoothing error:

$$
0.006394669\ldots;
$$

- theoretical bound:

$$
E_8
=
0.111007820\ldots;
$$

- quantized curvature range:

$$
0.22289999\ldots
\le
\widehat r
\le
0.77710000\ldots;
$$

- total support error:

$$
0.00946175\ldots;
$$

which is below the conservative theorem bound.

So the sanity test agrees with the analytic derivation.

---

# 33. Reproducibility Status

## UTF-8 canonical source

`PASS`

## Canonical math delimiters

Only:

$$
\$...\$
$$

and:

$$
\$\$...\$\$
$$

are used.

`PASS`

## Positive-kernel legality

`PROVED`

## Nonsmooth constant-width inclusion

`PROVED IN MEASURE FORM`

## Explicit kernel error

`PROVED`

## Legal coefficient quantization

`PROVED`

## Finite dictionary existence

`PROVED CONSTRUCTIVELY`

## Practical enumeration

`COMPUTE-DEFERRED`

## New Lebesgue numerical bound

`NONE`

## Global solution

`OPEN`

---

# 34. Round 03's Assigned Topic

Next round:

## AMRAL-LUC-FC-R03
### Signed Placement Margin and Full Fixed-Cover Finite Certificate

The goal is to genuinely compile:

$$
\min_{Q\in O(2)}
\min_{t\in\mathbb R^2}
\max_{\theta\in S^1}
$$

into a finite certified computation.

Priority tasks:

1. establish an exact angular sampling error for the support direction;
2. establish an exact rotation sampling error for the orientation;
3. handle translation exactly via convex LP / the active triple;
4. treat reflection as two discrete branches;
5. build a finite interval certificate schema;
6. merge Round 02's finite shape dictionary with the placement certificate.

If Round 03 succeeds, then for a **fixed candidate cover $U$**, we will for the first time obtain:

$$
\boxed{
\text{the entire target family}
+
\text{the entire congruent placement}
}
$$

finite-resolution rigorous certificate.

---

# 35. Shortest Handoff Conclusion

Round 02 completes:

$$
\boxed{
\mathcal R
\longrightarrow
\mathscr D_{N,\delta,q}
}
$$

and:

$$
\mathscr D_{N,\delta,q}
$$

is a genuinely finite, genuinely legal dictionary of unit constant-width bodies.

Total error:

$$
\boxed{
\varepsilon_{N,\delta,q}
=
\frac{
\pi^2(4N+1)
}{
2(2N^2+4N+3)(N+1)
}
+
\frac{\delta}{2}
+
(N-1)q.
}
$$

As long as:

$$
q
\le
\frac{
3\delta
}{
8(N-1)N(N+1)
},
$$

it is guaranteed that coefficient rounding does not break convexity.

Hence:

$$
\boxed{
\text{shape infinity has now been compressed into a certified finite-resolution domain.}
}
$$

The next obstruction has clearly moved to:

$$
\boxed{
\text{the placement continuum}.
}
$$

---

# References

1. J. C. Baez, K. Bagdasaryan, P. Gibbs, *The Lebesgue Universal Covering Problem*, arXiv:1502.01251, 2015.
2. P. Gibbs, *An Upper Bound for Lebesgue's Covering Problem*, arXiv:1810.10089, 2018.
3. U. Mishra, *Curves of constant width and Lebesgue's covering problem*, arXiv:2608.30538, 2026.
4. S. Zeng, *An exact hierarchy for Lebesgue's universal covering constant and a certified 0.834 lower bound*, arXiv:2609.01284, 2026.
5. N. K. Bari, *A Treatise on Trigonometric Series*, Pergamon, 1964.
6. A. Zygmund, *Trigonometric Series*, Cambridge University Press.
7. R. Shah, J. Urschel, N. West, *What is Jackson's constant?*, arXiv:2607.23375, 2026.
8. R. Schneider, *Convex Bodies: The Brunn–Minkowski Theory*, Cambridge University Press.
9. Neo.K + Aletheia, *AMRAL × Lebesgue Universal Covering — Round 01*, AMRAL-LUC-FC-R01 v0.1, 2026-09-18.

---

# 36. Declaration

This round does not claim to:

- solve the Lebesgue universal covering problem;
- obtain new strict upper or lower bounds on $a_{\mathrm{Leb}}$;
- automatically eliminate the placement continuum by means of the finite dictionary;
- replace the 2026 finite-certificate routes of Mishra or Zeng.

What this round has actually accomplished is:

$$
\boxed{
\text{a legality-preserving, explicit-error, finite-codebook constant-width shape compiler.}
}
$$

It is the foundational layer for the finite placement certificate and saturation audit that follow.