# Recursive SWC induction: the loop and the missing progress theorem

6 September 2026. **RH remains unproved.** This attempt investigates the suggestion to use recursive, looped induction to close the infinite-scale gap. It constructs a recursive approximation process and proves several of its properties for all iterations. The property that its error tends to zero remains open.

The subsequent [near-zero investigation](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/ALMOST_ZERO.md) proves explicit small-error bounds and a fixed-tolerance all-time SWC guarantee, while retaining the distinction from exact zero.

The finite contraction results through j = 10 in the [preceding report](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/LARGE_J.md) are preserved. No larger finite computation was needed for this investigation.

## 1. What a valid recursive induction would prove

The SWC source's *Well-founded reduction rule* requires a reduction to a smaller instance and a sound transformer from the smaller instance's witness. Thus recursion is already supported by the method. The requirement that matters here is a transformer whose validity is proved uniformly over the needed instances.

An approximation state can contain a finite coefficient vector, a rational upper error bound U > 0, and its exact verification record. Fix factors a_j in (0,1). A sufficient SWC loop contract is:

1. A verified base state exists.
2. For every admissible state at scale j, a specified generator has **proved positive mass** of outputs accepted as admissible states at scale j+1.
3. Every accepted output satisfies U_new ≤ a_j U_old.
4. The product of the a_j tends to zero.

Under these premises, SWC existence followed by ordinary induction supplies a state at each finite scale. The fourth premise makes their bounds tend to zero. The established approximation criterion then implies RH. A positive probability for one infinite run with no failures is not required for this logical argument: the induction establishes existence at each finite stage. A randomized implementation may retry failed proposals, but proving its next-stage success mass is part of premise 2.

For the previously proposed factors,

\[
 a_j=1-\frac1{2(j+1)},
 \qquad
 U_j\le U_J\sqrt{\frac{J+1}{j+1}}\longrightarrow0.
\]

The displayed bound follows from log(1−x) ≤ −x and a lower bound on the harmonic sum. Thus the implication from a valid loop contract to RH is proved. A universally valid next-state transformer with this progress property has **not** been constructed.

Recursive proof calls must discharge smaller obligations, or satisfy another proved sound progress condition. Reusing the desired next-step assertion inside its own justification does not discharge it. An invariant saying that every emitted certificate is correct also does not prove that the search will emit the next certificate.

## 2. An exact example: improvement forever can leave an error floor

This issue occurs inside the actual approximation system. Let g₂(x) = {1/(2x)}−{1/x}/2. It equals 1/2 on odd-indexed arithmetic cells and zero on even-indexed cells. If P₂ is projection onto span(g₂), then P₂1 = 2g₂ and

\[
 \|P_2 1\|_2^2=\log2,\qquad
 \|(I-P_2)1\|_2^2=1-\log2.
\]

Start r₀ = 1 and recursively set

\[
 r_{t+1}=(I-\tfrac12P_2)r_t.
\]

The two orthogonal components evolve independently, giving the exact formula

\[
 \boxed{\|r_t\|_2^2=1-\log2+(\log2)4^{-t}.} \tag{38}
\]

The error decreases strictly at every iteration, and induction proves that statement for every integer t. Its limit is nevertheless 1−log2 > 0. This is a counterexample to inferring convergence to zero from recursive strict improvement alone. It concerns a fixed one-dimensional approximation space, not a disproof of the full RH criterion.

## 3. A recursive loop using every integer basis direction

To avoid a fixed finite support, use the entire integer basis. Let H_cell be the closed subspace of L²(0,1) consisting of functions constant on each cell

\[
 I_m=(1/(m+1),1/m),\qquad w_m=|I_m|=1/[m(m+1)].
\]

The constant target χ = 1 and every g_k belong to this space. Working in H_cell is essential: on all of L²(0,1), functions with zero integral on every cell would give an irrelevant kernel automatically.

Normalize u_k = g_k/||g_k||₂, put q_k = 1/[k(k−1)] for k ≥ 2, and define

\[
 T=\sum_{k=2}^{\infty}q_k\,|u_k\rangle\langle u_k|.
\]

All q_k are positive rational numbers and their sum is 1. The series converges in trace norm. Hence T is a positive compact operator with trace 1 and 0 ≤ T ≤ I. Its kernel is

\[
 \ker T=\overline{\operatorname{span}\{g_k:k\ge2\}}^{\perp}
 \quad\text{inside }H_{\rm cell}.
\]

Indeed, ⟨Th,h⟩ is the sum of the nonnegative terms q_k|⟨h,u_k⟩|²; it vanishes exactly when all these inner products vanish.

The recursive process

\[
 r_0=\chi,\qquad r_{t+1}=(I-T)r_t \tag{39}
\]

has a fully specified update. If ν is the spectral measure of T for χ, then

\[
 \mathcal E_t:=\|r_t\|_2^2
 =\int_{[0,1]}(1-\lambda)^{2t}\,d\nu(\lambda),
 \qquad
 \lim_{t\to\infty}\mathcal E_t=\nu(\{0\}). \tag{40}
\]

These statements follow from the spectral theorem and dominated convergence. The discrete energy sequence has alternating nonnegative finite differences, since it is a moment sequence of (1−λ)² in [0,1]. In particular it is decreasing and convex. It decreases strictly at each finite iteration: χ has a nonzero component in the positive spectral subspace, and T has no eigenvalue 1. An eigenvector of eigenvalue 1 would have to be collinear with every u_k, contradicting independence of g₂ and g₃.

The remaining global statement is now precise:

\[
 \boxed{\nu(\{0\})=0.} \tag{41}
\]

This is equivalent to RH, because it says χ belongs to the closed span of the g_k. In this particular cell space it is also equivalent to ker T = {0}. To see the latter implication, note that integer dilation D_kf(x) = f(kx), with zero extension, preserves the closed span by

\[
 D_k g_n=g_{kn}-\frac1n g_k.
\]

If χ is in the span's closure, then so are all D_kχ and their differences, which are the cell indicators. Their span is dense in H_cell. The converse is immediate. This does not prove (41); it identifies its exact content.

### Finite realizations have a proved error bound

Let T_N be the sum through k = N. Its tail satisfies ||T−T_N|| ≤ 1/N. Therefore

\[
 \|(I-T)^t\chi-(I-T_N)^t\chi\|_2\le t/N. \tag{42}
\]

This is the telescoping bound for powers of two contractions. Moreover, χ−(I−T_N)^tχ is a finite combination of g₂,…,g_N. Thus every finite iteration admits a finite-dimensional realization, and (42) bounds the error caused by truncating the operator. Its Gram coefficients can in turn be enclosed or rationalized before using the existing SWC verifier.

Equations (39)–(42) provide a loop and a rigorous truncation rule. They establish convergence to the projection on the kernel. They do not establish that this projection is zero.

## 4. Positivity of every finite compression can be proved unconditionally

There is a useful exact calculation here. Let B_M be the M-by-M matrix

\[
 (B_M)_{m,k}=\frac{m\bmod k}{k},
 \quad 1\le m\le M,\quad2\le k\le M+1.
\]

Then

\[
 \boxed{\det B_M=\frac1{M+1}.} \tag{43}
\]

**Proof.** Replace each row after the first by its difference from the preceding original row. This determinant-one operation changes entry (m,k) to 1/k−1_(k|m). The last column is constantly 1/(M+1). Subtract (M+1)/k times this last column from each earlier column. Their first-row entries become zero; the remaining (M−1)-by-(M−1) block is the negative of a lower triangular divisibility matrix with diagonal 1. Expanding along the first row gives (43), including the signs. ∎

Let Q_M be projection onto the first M normalized cell indicators. Since ||g_k||₂² ≤ 1 and q_k ≥ 1/[M(M+1)] for 2 ≤ k ≤ M+1, the weighted basis matrix gives

\[
 \boxed{Q_MTQ_M\big|_{Q_MH_{\rm cell}}\ge\delta_M I_M,
 \qquad
 \delta_M=\frac1{M^M(M+1)^4(M!)^2}>0.} \tag{44}
\]

For completeness, write A_M = diag(√w_m)B_M. Its squared determinant is 1/[(M+1)^3(M!)²] and tr(A_MA_Mᵀ) ≤ M. Its smallest eigenvalue is consequently at least its determinant divided by M^(M−1). Multiplication by 1/[M(M+1)] yields (44).

Thus there is an unconditional positive-definiteness certificate for **every finite compression** of the actual operator. A nonzero kernel vector, if present, must have infinite cell support. The next section checks whether these finite certificates suffice to exclude it.

This does not violate SWC's pointwise rule: (44) certifies a family of finite-dimensional statements. The union of finitely supported vectors is dense but does not contain every vector in the infinite-dimensional space. Excluding a kernel on that union is a different conclusion from (41).

## 5. Exact countermodel: all finite compressions positive, limit error 9/25

This is an abstract counterexample to the proposed finite-to-infinite inference, not a counterexample to RH.

On ℓ² of the positive integers with standard basis e_n, define nonnegative unit vectors

\[
 v_n=\frac{e_n+2e_{n+1}}{\sqrt5},\qquad
 S=\sum_{n=1}^{\infty}2^{-n}|v_n\rangle\langle v_n|.
\]

This is a positive trace-class operator with trace 1, constructed from a full-support probability distribution over nonnegative vectors. Its matrix has entries

\[
 S_{11}=\frac1{10},\quad
 S_{ii}=\frac9{5\,2^i}\ (i\ge2),\quad
 S_{i,i+1}=S_{i+1,i}=\frac2{5\,2^i},
\]

and all other off-diagonal entries zero. The row sums are at most 3/4, so ||S|| ≤ 3/4.

For a vector supported in the first M coordinates, put x_(M+1) = 0. Then

\[
 \langle Sx,x\rangle
 =\frac15\sum_{n=1}^{M}2^{-n}|x_n+2x_{n+1}|^2
 \ge\frac{\|x\|^2}{5\,2^M(2^M-1)^2}. \tag{45}
\]

To justify the bound, the finite upper-bidiagonal map x ↦ (x_n+2x_(n+1)) has inverse norm at most 1+2+⋯+2^(M−1) = 2^M−1. Thus every finite compression is positive definite, with an explicit positive lower bound valid for every M. Its determinant is exactly

\[
 \det S_M=5^{-M}2^{-M(M+1)/2},
\]

by the same triangular factorization.

Nevertheless the normalized, infinitely supported vector

\[
 w_n=\sqrt3(-1)^n2^{-n}
\]

is in ker S, since w_n+2w_(n+1) = 0 for every n. These relations also show that ker S is exactly its one-dimensional span. Take the strictly positive normalized target f_n = √3·2^−n. Geometric summation gives

\[
 \langle f,w\rangle=-\frac35.
\]

The recursive loop f_(t+1) = (I−S)f_t therefore satisfies

\[
 \boxed{\lim_{t\to\infty}\|(I-S)^t f\|^2=\frac9{25}.} \tag{46}
\]

Its energy decreases strictly at every finite iteration: f has a nonzero component outside ker S, and all positive spectral values are below 1. Induction can prove continued improvement, all the finite matrices have positive certificates, and the limiting error is still positive.

Even requiring every matrix entry to be strictly positive does not rescue the inference. Replace S by

\[
 \widetilde S=\sum_{r\ge1}2^{-r}S^{2r}.
\]

The tridiagonal matrix S connects every pair of coordinates through paths, and its positive diagonal permits additional path steps; hence every entry of S̃ is positive. It has the same kernel as S, is positive and trace class, and every finite compression remains positive definite because S̃ ≥ S²/2. The limiting target component is again 9/25. This refinement still does not assert anything about the particular arithmetic operator T; it shows that these positivity properties alone cannot prove (41).

## 6. A rate bound rules out another shortcut for this loop

One might try to prove a simple polynomial decrease in the iteration number t for (39). For this choice q_k = 1/[k(k−1)], such a rate is impossible, even if RH is true.

Use the established [Burnol lower bound](https://arxiv.org/abs/math/0103058), which implies d_N² ≥ c/log N for some c > 0 and all sufficiently large N. Since χ−(I−T_N)^tχ belongs to H_N, equation (42) gives

\[
 \sqrt{\mathcal E_t}\ge d_N-t/N.
\]

Set N = t² for integer t. For large t, the term 1/t is at most half of the lower bound on d_N, so

\[
 \boxed{\mathcal E_t\ge\frac{c'}{\log t}}
 \quad\text{for some }c'>0\text{ and all sufficiently large }t. \tag{47}
\]

This is an unconditional bound on the chosen recursive loop. In particular, it excludes an upper bound O(t^−α) for any α > 0. The iteration counter t is different from the basis-size parameter j = log₂N in the previous attempts; their contraction formulas cannot simply be transferred by relabeling the counter.

It also excludes a power-law spectral estimate ν([0,ε]) ≤ Cε^α near zero for this T and any α > 0. Such an estimate would imply E_t = O(t^−α), by (1−λ)^(2t) ≤ exp(−2tλ) and integration by parts. A slower spectral estimate might still imply ν({0}) = 0, but none was established here.

## 7. A bounded inverse-energy argument is too strong for any fixed sampler

There is a further arithmetic obstruction to a common proposed progress bound. For a cell function F, let F_m denote its value on I_m and set F₀ = 0. For n ≥ 2 define the continuous finite-coordinate functional

\[
 \Lambda_n(F)=-\sum_{d\mid n}\mu(n/d)(F_d-F_{d-1}).
\]

The increment of g_k at cell d is 1/k−1_(k|d). Möbius inversion consequently gives

\[
 \Lambda_n(g_k)=\mathbf1_{n=k},\qquad
 \Lambda_n(\chi)=-\mu(n). \tag{48}
\]

These identities imply that any norm-convergent series χ = Σa_kg_k with fixed coefficients must have a_k = −μ(k). They do not prevent approximation by triangular coefficient arrays that readjust old coefficients as N grows, as our computations do.

Now allow **any** fixed full-support probability weights q_k, and construct T from the normalized g_k as before. There is no finite constant C satisfying

\[
 |\langle\chi,h\rangle|^2\le C\langle Th,h\rangle
 \quad\text{for every }h\in H_{\rm cell}. \tag{49}
\]

**Proof.** Define A:ℓ²→H_cell by Ae_k = √q_k g_k/||g_k||, so T = AA*. If (49) held, the map A*h ↦ ⟨χ,h⟩ would be a bounded linear functional. Extend it to the closure of this range and use the Hilbert-space representation theorem to obtain z ∈ ℓ² with χ = Az. The defining series for Az is absolutely convergent in Hilbert norm, since Σ√q_k|z_k| ≤ ||z||₂. Applying (48) termwise gives

\[
 z_k=-\mu(k)\frac{\|g_k\|_2}{\sqrt{q_k}}.
\]

On the cells with ceil(k/2) ≤ m < k, g_k ≥ 1/2 and their total length is at least 1/(2k). Hence ||g_k||₂² ≥ 1/(8k). In particular, since each q_k ≤ 1,

\[
 \sum_k|z_k|^2\ge\frac18\sum_{p\text{ prime}}\frac1p=\infty,
\]

contradicting z ∈ ℓ². The last divergence is Euler's classical theorem. ∎

Thus a bounded inverse-energy/source-condition proof of convergence would impose a condition that is actually false for this family of samplers. This does not rule out RH or SWC. It rules out that stronger proposed lemma and shows why a proof must allow weaker control near the zero end of the spectrum, or use a different adaptive construction.

In particular, this obstruction to a bound for every h does not refute a gain estimate restricted to a specially controlled sequence of approximation residuals. Such a sequence-specific estimate is still a possible proof target, but it has not been established.

## Status and reproduction

The new conclusions proved in this attempt are the conditional SWC loop contract, the all-iteration spectral identities, the unconditional finite-compression certificate (44), the countermodel (46), and the obstructions (47) and (49). **The assertion ν({0}) = 0 for the arithmetic operator T is not proved.** It is the remaining infinite-scale claim in this formulation.

This investigation therefore does not turn the existing finite evidence into an RH proof. It identifies which induction premise is missing and supplies exact counterexamples or bounds against several ways of trying to omit it.

The [exact checker](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/recursive_spectral.py) verifies the determinant and compression identities in twelve finite sizes and the geometric-series arithmetic for the 9/25 countermodel. The proofs for all sizes and all iterations are written above; the finite checks are not presented as substitutes for those arguments or as proof-assistant formalization. No novelty priority is claimed.

From `/Users/kt/Documents/ChatGPT/mathstuff`:

```bash
python3 research/swc-rh/recursive_spectral.py
python3 -m unittest discover -s research/swc-rh -p 'test_*.py' -v
```

The script uses the standard library and takes only a short time. Its [recorded results](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/recursive-spectral-checks.json) concern these new identities and the countermodel; the earlier large finite certificates were not regenerated.

The combined implementation suite now has **34 passing tests**, including the new coefficient-extraction identities and the interval estimate used in the inverse-energy obstruction.
