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The difference operator $Δ_{h} = Ψ_{h}^{*} - i d$ loses a controlled domain margin and is bounded by $C_{d} M ∣ h ∣ / b$

Statement

Let $d \geq 1$ and let $(U, U_{ρ}, \tilde{A}, \tilde{B}, M, K, δ, Ψ_{h})$ be analytic splitting data as in Analytic splitting data with uniform holomorphic neighbourhood, and the exact symmetric splitting map $Ψ_{h}$ , with complexified form and bracket as in Complex phase domains, holomorphic maps on $C^{m}$ , and the complexified canonical form and bracket. For $r > 0$ put $D (r) : = {ζ \in C^{2 d} : dist (ζ, \overline{U}) < r}$ .

There is a constant $C_{0} = C_{0} (d) \geq 1$ such that for every integer $N \geq 2$ , with

$b : = \frac{ρ}{8 N}, r_{N} : = \frac{ρ}{C_{0} M N}, q : = \frac{C_{d} M ∣ h ∣}{b},$

the following holds for every complex $h$ with $0 < ∣ h ∣ \leq r_{N}$ . Write $D_{k} : = D (ρ / 2 - 2 b k)$ for $0 \leq k \leq N$ ; then $D_{0} = D (ρ / 2)$ , $D_{N} = D (ρ / 4) \supseteq K$ , and $D_{N} \subseteq D_{k} \subseteq D_{0}$ for $k \leq N$ . Define the difference operator

$Δ_{h} f : = f \circ Ψ_{h} - f$

for functions $f$ holomorphic on $D_{k - 1}$ , where the composition is read on $D_{k}$ . Then:

Domain control. For every $k = 1, \dots, N$ the map $Ψ_{h}$ is holomorphic on $D_{k}$ , maps $D_{k}$ into $D_{k - 1}$ , and satisfies $∥ Ψ_{h} (Z) - Z ∥ \leq 2 M ∣ h ∣ \leq b$ for all $Z \in D_{k}$ .
Single-step bound. If $f$ is holomorphic on $D_{k - 1}$ with $∣ f ∣ \leq F$ there, then $Δ_{h} f$ is holomorphic on $D_{k}$ and $∣ Δ_{h} f (Z) ∣ \leq \frac{C_{d} M ∣ h ∣}{b} F (Z \in D_{k}),$ with $C_{d} : = 4 d^{2}$ .
Iterates. Taking $C_{0} : = \max {16, 128 d^{2}}$ one has $q \leq 1 / 4$ for $∣ h ∣ \leq r_{N}$ , and for every $f$ holomorphic on $D_{0}$ and every $1 \leq k \leq N$ , the iterate $Δ_{h}^{k} f$ is holomorphic on $D_{k}$ and $∣ Δ_{h}^{k} f (Z) ∣ \leq q^{k} ∣ f ∣_{D_{0}} (Z \in D_{k}) .$

Facts & Assumptions

Given: The analytic splitting data; $N \geq 2$ , $b = ρ / (8 N)$ , the domains $D_{k}$ ; the constant $C_{d}$ of clause 2 (to be produced in the proof).

[F1]

Geometric facts about the distance domains (Complex phase domains, holomorphic maps on $C^{m}$ , and the complexified canonical form and bracket): $dist (\cdot, \overline{U})$ is $1$ -Lipschitz, so $ζ \in D (r)$ implies $B (ζ, t) \subseteq D (r + t)$ for every $t > 0$ , and $D (r) \subseteq D (r^{'})$ for $r \leq r^{'}$ . Every $Z \in K \subseteq U$ has $dist (Z, \overline{U}) = 0$ , so $K \subseteq D_{N}$ .

[F2]

Flow displacement: by The flow of a bounded holomorphic field on a polydisc: existence in complex time, displacement bound, holomorphy, and the composition law clause 1, each of the three flows in $Ψ_{h} = φ_{\tilde{A}}^{h / 2} \circ φ_{\tilde{B}}^{h} \circ φ_{\tilde{A}}^{h / 2}$ displaces its argument by at most $M$ times the modulus of its time parameter, provided the argument lies in the appropriate polydisc of radius $s$ with $∣ t ∣ \leq s / (8 \cdot 2 d \cdot M)$ ; consequently, on the domains used below, $∥ Ψ_{h} (Z) - Z ∥ \leq (1 / 2 + 1 + 1 / 2) M ∣ h ∣ = 2 M ∣ h ∣$ . The composition is holomorphic by Hamiltonian flows and the exact splitting map preserve the complexified canonical form and the Poisson bracket and Calculus of holomorphic maps: chain rule, Wirtinger derivatives, coefficient extraction in one variable, and locally uniform limits clause 1.

[F3]

Cauchy derivative estimates on a polydisc (Iterated Cauchy formula, power-series expansion and Cauchy estimates for functions continuous and separately holomorphic on a polydisc clause 3): if $g$ is continuous and separately holomorphic on $Δ_{ρ} (a)$ , then for $0 < s_{k} < ρ_{k}$ , $∣ \partial_{z_{k}} g (a) ∣ \leq M (s) / s_{k}$ with $M (s) : = \sup_{Γ_{s} (a)} ∣ g ∣$ ; a function holomorphic on an open set is continuous and separately holomorphic there.

[F4]

The mean value estimate on a segment: if $g$ is totally differentiable on the segment $[Z, ξ]$ and $∣ \partial_{z_{k}} g ∣ \leq A$ on it for every $k$ , then $∣ g (ξ) - g (Z) ∣ \leq \sum_{k} A ∣ ξ_{k} - Z_{k} ∣ \leq \sqrt{2 d} A ∥ ξ - Z ∥$ ; this follows by applying the one-variable mean value estimate to the real coordinate functions along the segment (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ).

[F5]

Holomorphy of the pullback and of differences: $f \circ Ψ_{h}$ is holomorphic on $D_{k}$ by the chain rule of Calculus of holomorphic maps: chain rule, Wirtinger derivatives, coefficient extraction in one variable, and locally uniform limits clause 1, so $Δ_{h} f$ is holomorphic there (Complex phase domains, holomorphic maps on $C^{m}$ , and the complexified canonical form and bracket).

Proof

technique · direct; verify the margin bookkeeping, bound the increment by Cauchy estimates on slices of the segment, and iterate

1.1givenF1F2algebra

Margin bookkeeping. Fix $N \geq 2$ , $k \leq N$ and $Z \in D_{k}$ . Since $C_{0} \geq 128 d^{2}$ , $∣ h ∣ \leq r_{N}$ implies both $2 M ∣ h ∣ \leq b$ and $∣ h ∣ \leq ρ / (64 d \sqrt{2 d} M)$ . Construct the three factors sequentially using polydiscs of radius $s = ρ / (4 \sqrt{2 d})$ centred at their initial points. The first centre is $Z$ . The first flow exists by [F2] and displaces it by at most $M ∣ h ∣ / 2$ ; the second centre is therefore within that distance of $Z$ . Inductively, the second flow displaces its own initial point by at most $M ∣ h ∣$ , and the last by at most $M ∣ h ∣ / 2$ . Thus every intermediate centre and every point along a subflow, with its time parameter running along the segment from zero to that factor's time, lies within the accumulated bound $2 M ∣ h ∣ \leq b$ of $Z$ . At each centre the polydisc has Euclidean radius $ρ / 4$ , so its distance to $\overline{U}$ is less than $ρ / 2 + b + ρ / 4 \leq 13 ρ / 16 < ρ$ ; it lies in $U_{ρ}$ . The flow lemma permits each subtime because $s / (8 \cdot 2 d M) = ρ / (64 d \sqrt{2 d} M)$ . This induction establishes existence and the intermediate bounds without using cancellation in the final displacement. Set $ξ = Ψ_{h} (Z)$ ; the triangle inequality gives $∥ ξ - Z ∥ \leq 2 M ∣ h ∣ \leq b$ . By [F1], $dist (ξ, \overline{U}) < ρ / 2 - 2 b k + b = ρ / 2 - 2 b (k - 1) - b$ , so $ξ \in D_{k - 1}$ . The same strict bound holds along the straight segment $[Z, ξ]$ , since every such point is within $b$ of $Z$ . Consequently the polydisc of radius $b / \sqrt{2 d}$ about each point of that segment, including its distinguished boundary, lies in $D_{k - 1}$ . Composition holomorphy follows from [F2]. This proves clause 1 with the stated constants and margins.

2.1givenstep 1.1F3F4F5algebra

Single-step bound. Let $f$ be holomorphic on $D_{k - 1}$ with $∣ f ∣ \leq F$ . For $Z \in D_{k}$ , $ξ = Ψ_{h} (Z)$ and $ζ$ on the segment, [F3] applied at $ζ$ with the polyradius $(b / \sqrt{2 d}) 1$ , whose distinguished boundary lies in $D_{k - 1}$ by step 1.1, gives $∣ \partial_{z_{k}} f (ζ) ∣ \leq \sqrt{2 d} F / b$ for every $k$ . Hence by [F4], $∣ f (ξ) - f (Z) ∣ \leq \sqrt{2 d} \cdot \sqrt{2 d} \cdot \frac{F}{b} \cdot ∥ ξ - Z ∥ \leq \frac{2 d F}{b} \cdot 2 M ∣ h ∣,$ and $Δ_{h} f$ is holomorphic on $D_{k}$ by [F5]. This is clause 2 with $C_{d} : = 4 d^{2}$ , which depends only on $d$ .

3.1givenstep 1.1step 2.1algebra

Iteration. By induction on $k$ : suppose $Δ_{h}^{k - 1} f$ is holomorphic on $D_{k - 1}$ with $∣ Δ_{h}^{k - 1} f ∣ \leq q^{k - 1} ∣ f ∣_{D_{0}}$ (true for $k = 1$ ), then step 1.1 shows $Ψ_{h} (D_{k}) \subseteq D_{k - 1}$ , so $Δ_{h}^{k} f = Δ_{h} (Δ_{h}^{k - 1} f)$ is defined and holomorphic on $D_{k}$ by clause 2, with bound $q \cdot q^{k - 1} ∣ f ∣_{D_{0}} = q^{k} ∣ f ∣_{D_{0}}$ .

4.1givenstep 3.1algebra

The factor $q$ . With $C_{0} \geq 16$ the constraint $∣ h ∣ \leq r_{N}$ implies $2 M ∣ h ∣ \leq ρ / (8 N) = b$ and, with the constant $C_{d} = 4 d^{2}$ of step 2.1, $q = \frac{4 d^{2} M ∣ h ∣}{b} \leq \frac{4 d^{2} M ρ}{C_{0} M N} \cdot \frac{8 N}{ρ} = \frac{32 d^{2}}{C_{0}} \leq \frac{1}{4},$ because $C_{0} \geq 128 d^{2} = 32 C_{d}$ . This proves clause 3 with $q \leq 1 / 4$ .

5.1step 1.1step 2.1step 3.1step 4.1∎

Conclusion. Clause 1 is step 1.1, clause 2 is step 2.1, clause 3 is steps 3.1 and 4.1; all constants depend only on $d$ , $M$ , $ρ$ .

Remarks

The loss of margin is exactly $2 b = ρ / (4 N)$ per application: $N$ applications consume $ρ / 4$ of the available $ρ / 2$ , and the surviving domain $D (ρ / 4)$ still contains $K$ . The choices $b = ρ / (8 N)$ and $r_{N} = ρ / (C_{0} M N)$ are conventions fixed once and for all on this page; every later item refers to them.
The constant $q \leq 1 / 4$ is not used beyond the geometric summations; optimising it is irrelevant. No choice principle is used.

The difference operator $Δ_{h} = Ψ_{h}^{*} - i d$ loses a controlled domain margin and is bounded by $C_{d} M ∣ h ∣ / b$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The difference operator Δh=Ψh∗−idΔh​=Ψh∗​−id loses a controlled domain margin and is bounded by CdM∣h∣/bCd​M∣h∣/b

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The difference operator $Δ_{h} = Ψ_{h}^{*} - i d$ loses a controlled domain margin and is bounded by $C_{d} M ∣ h ∣ / b$