Theorem: Finite-time complete regulator

For fixed cutoffs, packets, and $T<\infty$, there is a finite prepared proxy isometry ${\hat{V}}_t$, embedded factorwise into the complete output, such that uniformly for $0\le t\le T$,

The same bound applies to both reduced channels and their diagnostic differences.

For $T>0$ and $0<b<1$, sufficient choices are

They give $b_{\mathrm{reg}}\le b$.

Proof. In the interaction picture,

On sectors with occupation at most $k$,

An $\ell$-fold product from vacuum has norm at most $(2G{)}^{\ell }\sqrt{\ell !}$. After integration over the ordered simplex, its Dyson term is bounded by $(2Gt{)}^{\ell }/\sqrt{\ell !}$. The series converges also after multiplication by any fixed polynomial in occupation. On these prepared vectors it solves the Schrödinger equation in the required domains and equals the self-adjoint evolution.

The compressed Hamiltonian $P_{m}H{P}_m$ has identical Dyson terms through order $m$. Inserted projections are contractions, so the two prepared isometries differ by at most $2Z_m(2GT)$.

Partition the frequency band into bins of width at most $\Delta$. Replacing frequency multiplication by bin midpoints gives $\Vert h-\tilde{h}\Vert \le \Delta /2$. On the occupation-cutoff space,

Both compressed Hamiltonians are bounded, so Duhamel's formula adds at most $Tm\Delta /2$.

The active one-particle space

is finite dimensional, invariant under $\tilde{h}$, and contains all couplings and collected packets. Its orthogonal complement remains vacuum. The total-occupation-cutoff space is finite dimensional and embeds into

This is an embedding, not an equality of the total cutoff with a tensor product. Source and carrier stay in the inaccessible factor. Lemma lem:stability converts the isometry allowance to (eq:breg).

Finally let $x=2GT$. The factorial estimate yields

Later term ratios are at most $1/2$, hence $Z_m(x)\le 2^{-m}$. The occupation and mesh terms are each at most $b/2$ under (eq:regulator-selection).