# Attempt to force the original minimum to zero using AI logic

Private working note. The attempt does not establish a zero minimum for the original RH approximation problem. It does establish a zero infimum for an enlarged class with a simple neural activation, and identifies the missing transfer back to the original class.

## 1. The exact stationary-residual question

Let S be the closed linear span of the original g_k in the arithmetic-cell Hilbert space. Let p be the orthogonal projection of the constant target 1 onto S and r=1-p. Then

    D = inf_{finite allowed F} ||1-F||_2^2 = ||r||_2^2,
    <r,g_k>=0 for every k.

For every permissible change v in S,

    ||1-(p+v)||_2^2 = ||r-v||_2^2 = D+||v||_2^2.

Thus if D>0, this residual is a genuine constrained global optimum. An AI optimizer restricted to the allowed directions cannot improve it merely by exploring more thoroughly. A proposed law that every positive residual admits an improving allowed direction would have to rule out this r; it is not a consequence of gradient descent.

There is a useful partial observation: if r>=0 almost everywhere, stationarity forces r=0. Indeed, a nonzero nonnegative cell function is positive on some cell I_m. Choosing k>m gives g_k=m/k>0 on that cell and g_k>=0 everywhere, so <r,g_k>>0, a contradiction. The same reasoning rules out a nonzero nonpositive stationary residual. Hence any hypothetical nonzero r must have both signs.

This suggests trying a neural activation that prevents overshoot and keeps the residual nonnegative. The following construction tests that route.

## 2. Clipping really does give a zero infimum in an enlarged class

Define, for integer k>=2,

    h_k(x)=min(1,k g_k(x)).

Because g_k>=0, this is the one-layer ReLU expression

    h_k = k g_k - ReLU(k g_k-1).

On arithmetic cell I_m, k g_k=m mod k is an integer. Consequently

    h_k = 0 when k divides m, and 1 otherwise.

Its squared error is exactly

    ||1-h_k||_2^2 = sum_{l>=1} 1/[k l(k l+1)]
                  < (1/k^2) sum_{l>=1} 1/l^2
                  < 2/k^2 -> 0.

The final bound uses sum l^-2 < 1+integral_1^infinity x^-2 dx=2. It is unconditional. For any positive tolerance, choosing a sufficiently large k produces a clipped approximant below that tolerance. One can give all such sufficiently large k positive sampling mass, or choose one deterministically; existence is already explicit.

**This proves zero infimum for the class containing the h_k. It does not prove zero infimum for the original finite linear span.** The activation is an additional operation whose preservation of the required class has not been established.

## 3. The activation fails to preserve the finite allowed class

For a cell function F, set F_0=0 and define, for n>=2,

    Lambda_n(F) = -sum_{d|n} mu(n/d)(F_d-F_{d-1}).

The earlier coefficient-extraction identity gives Lambda_n(g_k)=1 when n=k and zero otherwise. Therefore any finite combination F=sum_{k=2}^N c_k g_k satisfies Lambda_n(F)=0 for all n>N. These functionals are continuous because they involve finitely many cell values, each a bounded coordinate functional in the weighted Hilbert space.

Now consider the explicit clipped function h_3. For a prime p>3,

    Lambda_p(h_3)=h_3(1)-[h_3(p)-h_3(p-1)].

If p=2 mod 3, both p and p-1 are nonmultiples of 3, so the bracket is zero and Lambda_p(h_3)=1. There are infinitely many such primes: if a finite list contained them all, a prime divisor of 3 times their product minus 1 would supply a new prime congruent to 2 modulo 3. The constructed integer is not divisible by any listed prime, and an integer congruent to 2 modulo 3 must have at least one prime factor in that residue class.

Thus h_3 has nonzero coefficient extractions at arbitrarily large indices and cannot be any finite allowed combination. The new script checks the coefficient identity on a finite grid and the explicit h_3 values at several primes; the preceding argument, not the finite checks, proves the unbounded-index obstruction.

This does **not** prove h_3 lies outside S, the closed span. Infinite limits may belong to S without being finite combinations. Conflating these two statements would be another incorrect inference.

## 4. The remaining transfer is sufficient for RH

Suppose one could show h_k belongs to S for an unbounded sequence of k. Since h_k tends to 1 in norm and S is closed, this would imply 1 belongs to S and D=0.

More quantitatively, it suffices to construct finite allowed F_k with

    ||F_k-h_k||_2 -> 0,

because

    ||1-F_k||_2 <= sqrt(2)/k + ||h_k-F_k||_2 -> 0.

No such transfer family has been established in this attempt. Calling the activation an allowed AI reasoning operation does not establish its output's membership in S.

In fact, requiring all h_k to belong to S is an RH-equivalent family condition in this setup. One direction follows from the preceding norm limit. For the converse, if RH holds then 1 belongs to S. Integer dilation preserves S, since

    D_a g_k = g_(ak) - g_a/k,

where D_a f(x)=f(ax) with zero extension outside (0,1), and g_1=0. The differences D_m 1-D_(m+1) 1 are the cell indicators, whose span is dense in the cell space. Hence S is the whole cell space and contains every h_k. This uses the same [Báez-Duarte criterion](https://arxiv.org/abs/math/0202141) and [cell-space formulation](https://arxiv.org/abs/math/0607733) as the existing investigation.

There is also a direct quantitative way to see what an obstruction would do. If D>0 and r is the stationary residual above, then <r,1>=D. The clipped error bound implies

    |<r,h_k>-D| <= sqrt(2D)/k.

Since r is orthogonal to S, every F in S satisfies

    ||h_k-F||_2 >= |<r,h_k>|/sqrt(D)
                 >= sqrt(D)-sqrt(2)/k.

Thus the possible positive defect would reappear as an obstruction to approximating the neural outputs in the original class. The activation has not removed it.

## Outcome

The proposed AI operation achieves zero error in a modified approximation class. It does not establish that the original minimum is zero. The precise remaining target is a sound transfer of the clipped family back into the original closed span, or another argument that excludes the stationary residual r.

Run `python3 research/swc-rh/zero_minimum_attempt.py` for the finite exact checks. These include rigorous bounds on clipped-function errors, explicitly labeled as belonging to the enlarged class. They are not new RH certificates. No physical experiment, large optimizer run, publication, or update to the earlier PDF/ZIP was performed.
