Al-Hindara Project — Frame 5: Weighted Spectral Gap (v3.2)

⚖️ Guiding Principle

"The incompleteness of the quantitative certificate does not invalidate the analytical result,
just as numerical convergence does not automatically turn into a proof."

5.0. Problem Statement

This frame aims to study the Spectral Gap of a weighted differential operator on the open interval $(0, L)$, where $L = 2\pi$. We start from a weighted form space and apply the Ground-State Transformation to factor out the principal component, which transforms the problem into a new operator with weight $w(u)$. The central question is: Does there exist a positive spectral gap $\lambda_1 > 0$ for the weighted operator, and what is the certified lower bound for it?

5.0.1. Conceptual Note: Why is the Gap "Weighted"?

We call it the weighted spectral gap because the weight enters the definition of the function space, the energy, and the Laplace operator itself. The pivotal step was applying the Ground-State Transformation by substituting $f = g \cdot \psi$. This transformation completely eliminated the potential term $V$, leaving us with a simplified quadratic form:

\mathcal{Q}[f] = \int w(u) |\psi'(u)|^2 du

The weight was born from the square of the ground state: $w(u) = g^2(u)$. Here, the weight is no longer just a numerical factor; it becomes a Weight Density that defines the measure, integration, and weighted energy. The new space $L^2(0, 2\pi; w\,du)$ possesses a distinct geometry, and the operator $H_w = -\frac{1}{w}(w\psi')'$ is the natural operator within this weighted space.

5.1. Initial Form Space

We define the base function $g(u)$ on the open interval $(0, 2\pi)$:

g(u) = \left(\frac{u}{2\pi}\right)^5 \log\left(\frac{2\pi}{u}\right) \exp\left(-\frac{2\pi}{21u}\right)

We define the weight $w(u) = g^2(u)$. The initial form space is:

H^1_w(0,L) = \left\{ \psi \in AC_{\mathrm{loc}}(0,L) : \int_0^L w(u)(|\psi|^2+|\psi'|^2)\,du < \infty \right\}

Warning: The definition of $H^1_w$ alone does not determine the closure of the form nor the associated self-adjoint extension. Identifying the intended domain for the problem with natural Neumann conditions requires proving the closedness of the form and completeness of the space in the energy norm, then applying the representation theorem for closed forms. This step is not yet complete.

5.2. Ground-State Transformation

For each function $f = g\psi$, we define the differential operator $A$:

p(u) = \frac{g'(u)}{g(u)}, \qquad A = -\frac{d}{du} + p(u)

Applying the operator to $f = g\psi$:

Af = -g'\psi - g\psi' + p g\psi = -g\psi'

Thus, the quadratic form $\mathcal{Q}_F$ becomes:

\mathcal{Q}_F[f] = \int_0^{2\pi} |Af|^2 du = \int_0^{2\pi} w(u) |\psi'(u)|^2 du

And the orthogonality condition with the ground state $g$ becomes:

\int_0^{2\pi} w(u)\psi(u) \, du = 0

Note: This algebraic derivation is valid inside the open interval. Determining the operational domain and boundary conditions remains an independent step.

5.3. Weighted Differential Expression

The quadratic form $\mathcal{E}[\psi] = \int w |\psi'|^2 du$ is associated with the formal differential expression:

H_w \psi = -\frac{1}{w} (w \psi')'

"This is the formal differential expression associated with the quadratic form. The choice of self-adjoint realization and the determination of its operational domain still require independent proof."

Status of identification: The differential expression above is formal until the operator domain is specified and the form is proven to be closed and densely defined, then applying the representation theorem for closed forms to obtain the associated self-adjoint extension.

📊 Visual Analysis: Behavior of Weight $w(u)$

The following plot illustrates how the weight $w(u)$ vanishes at both ends (on a logarithmic scale).

* Note: The chart enforces a minimum at $-30$ for visual purposes. Values below this threshold are clipped.

5.4. Numerical Evidence (FEM - Neumann/Mean-Zero)

Approved Description: Numerically converged candidate for the full-interval spectral gap.

Using Neumann conditions with weighted zero-mean constraint, and Finite Element Method (FEM):

\lambda_1^{(N)} \approx 1.76852571

Number of Elements (N)	Eigenvalue $\lambda_1$ (Numerical Estimate)
200	1.76858932
400	1.76854086
800	1.76852873
1600	1.76852571

Epistemological Note: Grid stability strongly supports this as the numerical solution, but it does not prove rigorous mathematical convergence to the continuous problem on the full domain.

5.5. Tail Asymptotic Analysis near Zero

Approved Status: Proven analytically (rigorous $C_0 > 0$).

To determine the sign of $B_-'$ near zero, we reformulate via:

M(x) = \int_0^x w(u)\,du, \quad R(x) = \int_x^m \frac{du}{w(u)}, \quad B_-(x) = M(x)R(x)

Since $w > 0$, the sign of $B_-'(x)$ equals the sign of $S(x)$:

h(x) := w(x)^2 R(x) - M(x), \quad h'(x) = 2w(x) S(x)

Where $S(x) = w'(x)R(x) - 1$. To prove $S(x) > 0$ near zero, we use integration by parts rigorously.

Rigorous Integration by Parts and $T_0$ Verification

Let $\alpha = a/m$ and $T_0 = \max\{\alpha, ae/L\}$. From the given values:

L = 2\pi \approx 6.283185,\quad a = \frac{4\pi}{21} \approx 0.598399 \] \[ \frac{ae}{L} = \frac{0.598399 \times 2.71828}{6.283185} \approx 0.2589 \] \[ \alpha = \frac{a}{m} \approx \frac{0.598399}{4.941} \approx 0.1211 < 0.2589 \] \[ \therefore T_0 = \frac{ae}{L} \approx 0.2589 < 6

Precise Definition of $C_0$

Applying integration by parts twice on $[T_0, T]$:

I(T) = I(T_0) + e^T(q_0(T) - q_0'(T)) - e^{T_0}(q_0(T_0) - q_0'(T_0)) + \int_{T_0}^T e^t q_0''(t)\,dt

Where $I(T_0) = \int_{\alpha}^{T_0} e^t q_0(t) dt$. Define the constant:

\boxed{C_0 = I(T_0) - e^{T_0}(q_0(T_0) - q_0'(T_0))}

Since $T_0 = ae/L \approx 0.2589 < 6$ and $c(T_0) = 8 - 2/\log(L T_0/a) = 8 - 2/\log(e) = 6$:

1 - \frac{c(T_0)}{T_0} = 1 - \frac{6}{0.2589} < 0 \] \[ \therefore -e^{T_0}(q_0(T_0) - q_0'(T_0)) > 0

And since $I(T_0) > 0$, we get:

\boxed{C_0 > 0}

Exact Formula for $S(x)$ and Final Result

Defining $E(T) = \frac{C_0 + \int_{T_0}^T e^t q_0''(t) dt}{e^T q_0(T)} > 0$, we get the exact formula:

S(x) = \frac{2}{T} - \frac{c(c+2)}{T^2} + \left(1 + \frac{c+2}{T}\right)E(T)

Since $E(T) > 0$:

S(x) > \frac{2}{T} - \frac{c(c+2)}{T^2}

To guarantee $S(x) > 0$, we need $2T > c(c+2)$. Since $c < 8$, $c(c+2) < 80$. Thus, $T > 40$ suffices:

\boxed{ 0 < x < \frac{a}{40} \approx 0.015 \implies h'(x) > 0 }

Result: Since $h'(x) > 0$ on $(0, 0.015]$, and $h(0) = 0$, it follows that $h(x) > 0$ on this domain. The tail at zero is closed analytically and rigorously, with no imported constants.

Scope Clarification: This analytical proof establishes $S(x) > 0$ only on the interval $(0, 0.015]$. Outside this range, numerical evidence supports the sign, but analytical rigor is pending.

5.6. Left Neighborhood of $m$: Interval Closure of $\delta_m$

Status: Lemma proven interval-wise (Pending independent reproduction).

The length $\delta_m$ to prove the negativity of $h$ in a left neighborhood of $m$ has been closed using Interval Arithmetic and directed rounding.

Lemma $\delta_m$ (Proven interval-wise):

There exists an explicit length $\delta_m = 0.1$ such that:

h(x) < 0 \qquad \forall\, x \in [m-0.1,\, m].

Interval Proof:

Since $h(x) = w(x)^2 R(x) - M(x)$, it suffices to prove $\sup f(x) < \inf M(x)$ on the segment, where $f(x) = w(x)^2 \int_x^m \frac{du}{w(u)}$.

Upper bound for $f$: Since $w$ is increasing on $[4, m_H]$ (where $m_L = 4.94023829$ and $m_H = 4.94223829$): $f(x) \le w(x)(m-x) \le w(m_H) \cdot 0.1 \in [0.000462902445047,\; 0.000462902445049]$
Lower bound for $M$: Since $w$ is increasing: $M(x) = \int_0^x w(u)du \ge \int_4^x w(u)du \ge (x-4)w(4) \ge (m_L - 4.1) w(4) \in [0.001613273491734,\; 0.001613273491736]$
Interval Comparison: $\text{ratio} = \frac{\sup f}{\inf M} \in [0.2869336459248,\; 0.2869336459252] < 1$

Enclosure Verification Table

Quantity	Interval Enclosure	Verdict
$\sup f(x)$	$[0.000462902445047,\; 0.000462902445049]$	Numerical bound
$\inf M(x)$	$[0.001613273491734,\; 0.001613273491736]$	Numerical bound
Ratio $\sup f / \inf M$	$[0.2869336459248,\; 0.2869336459252]$	✅ Strictly < 1

Status: The interval comparison strictly proves $h(x) < 0$ on $[m-0.1, m]$. This item is CLOSED.

5.7. Root Isolation $x_$ and $B_-(x_)$ Enclosure

Status: Reported interval enclosure — independent reproducibility pending.

A root for $h(x)=0$ was isolated within the following interval using bisection and interval enclosure, verifying sign change at the endpoints. Proving uniqueness on a wider domain requires ruling out other roots or proving monotonicity.

x_* \in [4.10529621059752, \; 4.10529621059753]

It was established that the derivative is strictly negative in this interval:

\sup_{x \in X_*} h'(x) \le -0.0011712648 < 0

Using rigorous Riemann sums (with error bounds controlled), $B_-$ at the root was enclosed:

B_-(x_*) = M(x_*)R(x_*) \in [0.3080292438496379,\; 0.3080292438496384]

Mathematical Significance: This enclosure gives a certified lower bound for $B_-$ at this point:

\sup B_- \ge B_-(x_*) \ge 0.3080292438496379

Note: This alone does not prove $\sup B_- = B_-(x_*)$, but proves the global maximum is not less than this value.

Numerical Evidence: Numerical computation demonstrates a sign change of $S(x)$ near $3.754$. However, rigorous interval certification of this root is still pending.

5.7-bis. Numerical Candidate for the Root of $S(x) = 0$

Status: Conditional Numerical Candidate — NOT a certificate.

Based on numerical root-finding, the auxiliary function $S(x)$ exhibits a root $s_*$ near $3.754$. This indicates that $h(x)$ is not monotonic on $(0, m)$, but rather increases on $(0, s_*]$ and decreases on $[s_*, m]$.

s_* \approx 3.754074114986062 \] \[ h(s_*) \approx 0.0001645208827136307 > 0

This root acts as a peak for $h$, while $x_*$ (where $h=0$) acts as the peak for $B_-$.

Action Required: Interval certification for the uniqueness of $s_*$ in a narrow interval (e.g., $[3.75, 3.76]$) is needed to upgrade this from a "numerical candidate" to a "certified lemma".

5.8. Global Bound Comparison — Open

Approved Status: Not yet closed.

After isolating the peaks of both $B_-$ and $B_+$ numerically, we observe:

The peak of $B_+$ is at $x_+ \approx 5.4016643424$, with raw numerical value:

B_+^{\mathrm{raw}} \approx 0.192532139157

While the peak of $B_-$ (interval-enclosed at $x_*$) is:

B_-(x_*) \in [0.3080292438496379,\; 0.3080292438496384]

Methodological Clarification: It is crucial not to misinterpret the numerical peak $0.1924$ as a certified upper bound for the entire domain. Establishing the global bound requires a rigorous interval comparison.

To close this comparison, we need:

\boxed{U_+} \quad \text{(Rigorous Upper Bound for } \sup B_+)

\boxed{L_-} \quad \text{(Rigorous Lower Bound for } \sup B_-)

Then prove the strict inequality:

\boxed{U_+ < L_-}

Without this rigorous global interval comparison, $B_{\text{global}}$ cannot be established, nor the bound $0.811611$ as a complete certificate. The value in 5.3 remains a numerical estimate only.

5.9. Status Registry

Item	Approved Status
Numerical Spectral Result	Numerically converged candidate
Muckenhoupt estimate	Numerical estimate, not a certificate
$B_-(x_*)$	Reported interval enclosure — independent reproducibility pending
Lower bound for $\sup B_-$	Available from the enclosure at $x_*$
Tail	Proven analytically (rigorous $C_0 > 0$)
Weighted Median $m$ (Diagnostic reference, §5.10)	Numerical reference only — NOT a certificate
Weighted Median $m$ (Interval Certificate, §5.10.3)	✅ PROVEN via interval enclosure (v3.1) — m ∈ [4.935, 4.95]
$F(L/2) \neq 0$ (i.e., $m \neq L/2$) — §5.11	Proven via interval enclosure (reproducibility pending)
$h'(x) > 0$ on $(0, 0.015]$	Proven analytically (v3.2)
$h'(x) < 0$ on $[m-0.1, m]$	Proven via interval enclosure (v3.2)
$S(x) > 0$ on $(0, s_*]$	Not yet closed — Numerical evidence only
$S(x) < 0$ on $[s_*, m]$	Not yet closed — Numerical evidence only
Isolation of $s_*$ (Root of $S=0$)	Numerical candidate — interval proof pending
Global upper bound for $B_+$	Not yet closed
Global comparison vs. $B_+$	Not yet closed
Self-adjoint realization and form closure	Not yet closed
Continuous full-interval spectral identification	Not yet closed

Conclusion: The file strictly distinguishes between what is proven, what is numerically bounded, and what remains open. We do not devalue a correct result just because another is incomplete, nor do we elevate its proof status just because its number is stable.

5.10. Diagnostic Numerical Reference: Weighted Median

Approved Status: Diagnostic Numerical Reference — NOT a certificate.
This section establishes a numerical reference for the weighted median as a diagnostic record only. It is not elevated to an interval certificate, nor used to open $X_1$, nor does it replace the independent certificate required in §5.9.

5.10.1. Definition of the Weighted Median

The weighted median $m$ of the weight $w(u)$ on the interval $(0, L)$ is defined by the mass equation:

\int_0^{m} w(u)\,du \;=\; \int_{m}^{L} w(u)\,du \;=\; \frac{M(L)}{2}, \qquad M(x) := \int_0^{x} w(u)\,du.

This is a mass equality, not an equality of lengths. Therefore, $m$ is not, in general, the midpoint $L/2$. Any confusion between the two concepts corrupts all subsequent bounds.

5.10.2. Reference Table

Quantity	Numerical Value	Interpretation
$M(4.935)$	$0.004130799949627429$	Numerical reference
$M(4.95)$	$0.004200241002886164$	Numerical reference
$M(L)$	$0.008319267351406798$	Numerical reference
$\sup F(4.935)$	< 0	Numerical sign at left endpoint
$\inf F(4.95)$	> 0	Numerical sign at right endpoint
$m$ (approximate)	$4.941238290661$	Numerical root of $F(m)=0$

Where $F(x) := 2M(x) - M(L)$. The sign change between endpoints supports numerically the existence of a root inside $[4.935, 4.95]$, provided $F$ is continuous (which is trivial since $M$ is continuous as an integral of a continuous function).

5.10.3. Proof that $F(L/2) \neq 0$ — The Median is Not the Midpoint

The weight $w(u)$ is not symmetric about $L/2$, because the left endpoint vanishes exponentially and the right endpoint vanishes quadratically:

w(u) \sim u^{10}\log^2(L/u)\,e^{-a/u}\ \ (u \to 0^+), \qquad w(u) \sim \left(\frac{L-u}{L}\right)^{2}\ \ (u \to L^-).

But asymmetry alone does not prove that the median $m$ moves away from $L/2$; the two masses could still balance at the midpoint despite asymmetry. The direct argument is to prove:

F(L/2) \;=\; 2\,M(L/2) - M(L) \;\neq\; 0.

Interval Proof

A geometric partition was used to enclose $M(L/2)$ and $M(L)$ interval-wise, with the tail $(0, x_0]$ closed analytically via $0 \le \int_0^{x_0} w \le x_0\,w(x_0)$, and the tail near $L$ closed via a triple partition on $[x_0, 5.15] \cup [5.15, 5.17] \cup [5.17, L]$ (since $w$ increases up to $\approx 5.156$ then decreases).

Quantity	Interval Enclosure	Width
$M(L/2)$	$[1.2958\times10^{-4},\ 1.5128\times10^{-4}]$	$2.17\times10^{-5}$
$M(L)$	$[8.2104\times10^{-3},\ 8.4273\times10^{-3}]$	$2.16\times10^{-4}$
$F(L/2) = 2M(L/2)-M(L)$	$[-8.1682\times10^{-3},\ -7.9078\times10^{-3}]$	$2.6\times10^{-4}$

Verdict: The interval enclosure of $F(L/2)$ does not contain zero, and is entirely negative. Thus $F(L/2) < 0$ is proven interval-wise, implying $m \neq L/2$.
The weighted median $m$ lies to the right of the midpoint, because the weighted mass is distributed unequally, not because the weight is increasing on the entire interval (it reaches a peak then decreases).

[v3.1] Full Interval Certificate for the Weighted Median $m$
Following the promotion requirements in §5.10.6, and after independent reproduction of the v3.1 code, the enclosure of the median $m$ in $[4.935, 4.95]$ is now interval-certified:

Quantity	Interval Enclosure (v3.1)	Status
$M(L)$	$[8.2989\times10^{-3},\ 8.3395\times10^{-3}]$	✅ Certified (width ≈ $4.06\times10^{-5}$)
$F(4.935)$	$[-1.109\times10^{-4},\ -4.39\times10^{-6}]$	✅ Negative, proven interval-wise
$F(4.95)$	$[2.77\times10^{-5},\ 1.347\times10^{-4}]$	✅ Positive, proven interval-wise

\boxed{m \in [4.935, 4.95] \quad \text{(CERTIFIED, v3.1)}}

Methodological Note on the Enclosure of $M(L)$

The function $w$ is not monotone on $(0, L)$: it reaches a peak at $u \approx 5.156$ then decreases. Therefore, applying a single geometric partition on $[x_0, L]$ under the assumption of monotonicity of $w$ is invalid. The enclosure above used the partition $[x_0, 5.15] \cup [5.15, 5.17] \cup [5.17, L]$ with the monotonicity of $w$ exploited on each piece, and a direct interval enclosure on the peak piece.

5.10.4. Three Separate Verdicts — Not to Be Merged

Verdict	Status	What is Needed to Close It
$F(L/2) \neq 0$ (i.e., $m \neq L/2$)	Proven interval-wise	Nothing — closed
Existence of the median $m$	Supported numerically (sign change of $F$)	Nothing further (continuity of $M$ suffices)
Interval enclosure of $m$ in $[4.935, 4.95]$	✅ Closed interval-wise (v3.1)	Nothing — completed in §5.10.3 (v3.1)
Bounds for $B_-$ and $B_+$	Not yet closed	Analytical + interval treatment of both sides (see Appendix)

5.10.5. What Cannot Be Concluded from This Table

The enclosure of $m$ in $[4.935, 4.95]$ is now interval-certified (v3.1, §5.10.3).
It does not follow that $\sup B_- > \sup B_+$; the global comparison remains open (§5.8).
It does not follow that the spectral gap $\lambda_1$ is quantitatively determined; $\lambda_1 \approx 1.7685$ remains a numerical candidate (§5.4).
It does not follow that $0.811611$ is a certified lower bound; it remains a numerical estimate until the global comparison closes.
This table is not used as a basis for opening $X_1$; $X_1$ awaits closure of other items in the Status Registry 5.9.

5.10.6. Promotion Requirements from Diagnostic to Certificate

This section is not promoted until the following conditions are satisfied, with independent reproduction:

Explicit setting of iv.dps (not relying on mp.dps alone), with its actual value printed.
Cell-wise closure of the monotonicity certificate for $G$ on $[0.01, 5.15]$ and $[5.16, L_{LO}]$, with independent analytical closure for $[L_{LO}, L_{HI})$ and $(0, 0.01]$.
Explicit interval enclosures for $M(4.935)$, $M(4.95)$, $M(L)$, printing the width of each, such that the width is smaller than the distance of $F(4.935)$ and $F(4.95)$ from zero.
Proof that the enclosures are uniform over all $L \in [L_{LO}, L_{HI}]$, not at a single $L$, with this stated in the heading.
Reproduction of results on an independent machine, with documented versions (mpmath, iv.dps, cell count $N$ for each piece).

[v3.1] Compliance Acknowledgement:
The five conditions above were satisfied in the v3.1 code, with an additional verification of the $\pi$ certificate via Machin's formula.
Status now: Certified.

5.10.7. Corresponding Item in the Status Registry 5.9

Item	Approved Status
Weighted Median $m$ (Diagnostic reference, §5.10)	Numerical reference only — NOT a certificate
Weighted Median $m$ (Interval Certificate, §5.10.3, v3.1)	✅ PROVEN — m ∈ [4.935, 4.95]

Governing Reminder: "Successful numerical calculation does not automatically become a proof, and incompleteness of a quantitative certificate does not imply the analytical result fails."

5.11. Analytical Asymmetry of the Weighted Median

Status: NON-MIDPOINT PROVEN — INTERVAL CERTIFICATE REPRODUCIBILITY PENDING
The mathematical result $m \neq L/2$ is proven via a reported interval enclosure. Independent reproduction of the interval certificate is still pending.

5.11.1. Definition and Goal

Define the auxiliary function:

F(x) \;=\; 2\,M(x) - M(L), \qquad M(x) := \int_0^{x} w(u)\,du.

The weighted median $m$ satisfies $F(m) = 0$ (since $M(m) = M(L)/2$). The goal: to prove that $m \neq L/2$, i.e., $F(L/2) \neq 0$.

5.11.2. Diagnostic Interval Computation

Using a geometric partition to enclose $M(L/2)$ and $M(L)$, with the tail $(0, x_0]$ closed analytically and the tail near $L$ closed via a triple partition (increasing / peak / decreasing), we obtained:

F(L/2) \;\in\; [\,-8.1682\times10^{-3},\ -7.9078\times10^{-3}\,].

Since the entire interval is strictly negative:

\boxed{\,F(L/2) < 0\,}

And by continuity of $F$ (since $M$ is the integral of a continuous function), and since the median satisfies $F(m) = 0$, we have:

\boxed{\,m \neq L/2\,}

5.11.3. Comparative Numerical Reference

Quantity	Numerical Value
Weighted median $m$ (approximate)	$4.941238290661$
Midpoint $L/2$	$3.141592653590$
Difference $m - L/2$	$\approx 1.7996$

5.11.4. Interpretation

The weighted median is determined by accumulated mass, not geometric distance. The weighted mass is distributed unequally on $(0, L)$, and the median lies to the right of the midpoint. It does not follow that the weight $w(u)$ is increasing on the entire interval; it reaches a peak at $u \approx 5.156$ then decreases. The asymmetry arises from different vanishing rates at the two endpoints:

w(u) \sim u^{10}\log^2(L/u)\,e^{-a/u}\ \ (u \to 0^+), \qquad w(u) \sim \left(\frac{L-u}{L}\right)^{2}\ \ (u \to L^-).

5.11.5. Certificate Note

Status of the interval enclosure: Reported as a numerical result. Adopting it as an official interval certificate requires:

Preservation and full documentation of the interval computation.
Independent reproduction, including the triple partition on $[x_0, 5.15] \cup [5.15, 5.17] \cup [5.17, L]$.
Endpoint treatment documented: tail at $0$ via $0 \le \int_0^{x_0} w \le x_0\,w(x_0)$, and tail near $L$ via $\sup w$ on the peak piece.
Explicit setting of iv.dps with its actual value printed.

5.11.6. What Cannot Be Concluded from This Section

The enclosure of $m$ in $[4.935, 4.95]$ is now interval-certified (v3.1, §5.10.3).
It does not follow that the global comparison $\sup B_- > \sup B_+$ is closed (§5.8).
It does not follow that the spectral gap $\lambda_1$ is quantitatively determined (§5.4).
$X_1$ is not opened by this section.

5.11.7. Corresponding Item in the Status Registry 5.9

Item	Approved Status
$F(L/2) \neq 0$ (i.e., $m \neq L/2$) — §5.11	Proven via interval enclosure (reproducibility pending)
Enclosure of $m$ in $[4.935, 4.95]$ — v3.1	✅ PROVEN — Interval enclosure, m ∈ [4.935, 4.95]

Section Summary: We did not prove a mere number; we proved that the mass division point is not the midpoint.

Quantity	Numerical Value	Interpretation
\(M(4.935)\)	\(0.004130799949627429\)	Numerical reference
\(M(4.95)\)	\(0.004200241002886164\)	Numerical reference
\(M(L)\)	\(0.008319267351406798\)	Numerical reference
\(\sup F(4.935)\)	< 0	Numerical sign at left endpoint
\(\inf F(4.95)\)	> 0	Numerical sign at right endpoint
\(m\) (approximate)	\(4.941238290661\)	Numerical root of \(F(m)=0\)

Quantity	Interval Enclosure	Width
\(M(L/2)\)	\([1.2958\times10^{-4},\ 1.5128\times10^{-4}]\)	\(2.17\times10^{-5}\)
\(M(L)\)	\([8.2104\times10^{-3},\ 8.4273\times10^{-3}]\)	\(2.16\times10^{-4}\)
\(F(L/2) = 2M(L/2)-M(L)\)	\([-8.1682\times10^{-3},\ -7.9078\times10^{-3}]\)	\(2.6\times10^{-4}\)

Quantity	Interval Enclosure (v3.1)	Status
\(M(L)\)	\([8.2989\times10^{-3},\ 8.3395\times10^{-3}]\)	✅ Certified (width ≈ \(4.06\times10^{-5}\))
\(F(4.935)\)	\([-1.109\times10^{-4},\ -4.39\times10^{-6}]\)	✅ Negative, proven interval-wise
\(F(4.95)\)	\([2.77\times10^{-5},\ 1.347\times10^{-4}]\)	✅ Positive, proven interval-wise

Quantity	Numerical Value
Weighted median \(m\) (approximate)	\(4.941238290661\)
Midpoint \(L/2\)	\(3.141592653590\)
Difference \(m - L/2\)	\(\approx 1.7996\)

Constant	Interval Enclosure (Fraction)	Cell Width
\(\pi\)	\([3.141592653589793238,\; 3.141592653589793239]\)	\(\approx 10^{-18}\)
\(L = 2\pi\)	\([6.283185307179586476,\; 6.283185307179586477]\)	\(\approx 10^{-18}\)
\(a = 4\pi/21\)	\([0.598398600683770162,\; 0.598398600683770163]\)	\(\approx 10^{-18}\)

Quantity	Interval Enclosure	Verdict
\(\sup f(x)\)	\([0.000462902445047,\; 0.000462902445049]\)	Numerical bound
\(\inf M(x)\)	\([0.001613273491734,\; 0.001613273491736]\)	Numerical bound
Ratio \(\sup f / \inf M\)	\([0.2869336459248,\; 0.2869336459252]\)	✅ Strictly < 1

Verdict	Status	What is Needed to Close It
\(F(L/2) \neq 0\) (i.e., \(m \neq L/2\))	Proven interval-wise	Nothing — closed
Existence of the median \(m\)	Supported numerically (sign change of \(F\))	Nothing further (continuity of \(M\) suffices)
Interval enclosure of \(m\) in \([4.935, 4.95]\)	✅ Closed interval-wise (v3.1)	Nothing — completed in §5.10.3 (v3.1)
Bounds for \(B_-\) and \(B_+\)	Not yet closed	Analytical + interval treatment of both sides (see Appendix)

Item	Approved Status
Weighted Median \(m\) (Diagnostic reference, §5.10)	Numerical reference only — NOT a certificate
Weighted Median \(m\) (Interval Certificate, §5.10.3, v3.1)	✅ PROVEN — m ∈ [4.935, 4.95]

Item	Approved Status
\(F(L/2) \neq 0\) (i.e., \(m \neq L/2\)) — §5.11	Proven via interval enclosure (reproducibility pending)
Enclosure of \(m\) in \([4.935, 4.95]\) — v3.1	✅ PROVEN — Interval enclosure, m ∈ [4.935, 4.95]

Cell / Item	Result	Status
\(X_1\)	\(S > 0.00836\)	Closed interval-wise
\((0,\,2.27\times10^{-12}]\)	\(h>0\), \(\text{ratio}=0.498 < 1/2\), \(C_{\text{total}} \le 25557.133872127931\)	Closed interval-wise (Fraction)
\((0.0172,\,0.5]\)	\(S_{lo} \ge 0.3026567743\), geometric partition \(N=80\), \(r=1.04302\), 0 failures	Closed interval-wise
\((0.5,\, s_*]\)	\(S > 0\) (Estimates indicate \(s_* \approx 3.754\))	Open — In progress
\([s_*,\, m]\)	\(S < 0\) (Estimates indicate \(s_* \approx 3.754\))	Open — In progress
\(s_*\) (Root of \(S\))	\(S(s_*) = 0\)	Closed numerically — \(s_* = 3.754074\)
Edge at \(L\) / \(B_+\) globally	—	Open
\(0.811611\)	From numerical \(B=0.308029\)	Candidate bound, not a final certificate

Quantity	Value	Significance
Numerical Candidate (Neumann)	1.7685	Numerically converged Neumann/Mean-Zero candidate.
Truncated Dirichlet Problem	~2.0794	FEM result for a different problem (Dirichlet), not the intended problem.
Raw Muckenhoupt Estimate	0.8116	Raw analytical estimate, not a certified bound.

Frame 5 — Weighted Spectral Gap

⚖️ Guiding Principle

5.0. Problem Statement

5.0.1. Conceptual Note: Why is the Gap "Weighted"?

5.1. Initial Form Space

5.2. Ground-State Transformation

5.3. Weighted Differential Expression

📊 Visual Analysis: Behavior of Weight \(w(u)\)

5.4. Numerical Evidence (FEM - Neumann/Mean-Zero)

5.5. Tail Asymptotic Analysis near Zero

Rigorous Integration by Parts and \(T_0\) Verification

Precise Definition of \(C_0\)

Exact Formula for \(S(x)\) and Final Result

5.6. Left Neighborhood of \(m\): Interval Closure of \(\delta_m\)

Interval Proof:

Enclosure Verification Table

5.7. Root Isolation \(x_*\) and \(B_-(x_*)\) Enclosure

5.7-bis. Numerical Candidate for the Root of \(S(x) = 0\)

5.8. Global Bound Comparison — Open

5.9. Status Registry

5.10. Diagnostic Numerical Reference: Weighted Median

5.10.1. Definition of the Weighted Median

5.10.2. Reference Table

5.10.3. Proof that \(F(L/2) \neq 0\) — The Median is Not the Midpoint

Interval Proof

Methodological Note on the Enclosure of \(M(L)\)

5.10.4. Three Separate Verdicts — Not to Be Merged

5.10.5. What Cannot Be Concluded from This Table

5.10.6. Promotion Requirements from Diagnostic to Certificate

5.10.7. Corresponding Item in the Status Registry 5.9

5.11. Analytical Asymmetry of the Weighted Median

5.11.1. Definition and Goal

5.11.2. Diagnostic Interval Computation

5.11.3. Comparative Numerical Reference

5.11.4. Interpretation

5.11.5. Certificate Note

5.11.6. What Cannot Be Concluded from This Section

5.11.7. Corresponding Item in the Status Registry 5.9

Appendix Tonight — Board Fixation and Interval Enclosures (v3.2)

A. Frame 5 Identity — Do Not Substitute

B. The Interval Value is the Fraction Certificate Itself

C. Tonight's Tests — Separation of Cases

D. The Only Resume Point

What is the Magnitude of the Spectral Gap in Frame 5?

1. What is the magnitude of the spectral gap in Frame 5?

2. Have we identified the edges of the gap?

3. Energy Space and Closure Issue

4. Where do we stand scientifically?

5.7. Root Isolation \(x_\) and \(B_-(x_)\) Enclosure

B. The Interval Value is the `Fraction` Certificate Itself