CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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A small angular cone can accumulate order-one phase error

Physical scope

Counterexample inside the ideal band-limited scalar angular-spectrum model in a fixed inertial chart and SI units. Its spectra are square-integrable plane data and are not asserted to be outputs of a compact finite aperture. It is a conditional model calculation, not an observed optical result.

Statement refuted

For every sequence $(δ_{n})_{n \geq 1}$ with $δ_{n} \in (0, 1)$ and $δ_{n} \to 0$ , every fixed $k_{0} > 0$ , every nonzero scalar input $u_{0, n} \in L^{2} (R^{2}; C)$ whose Fourier support lies in $∣ 2 π ξ ∣ \leq δ_{n} k_{0}$ , and every sequence of forward distances $z_{n} > 0$ satisfying $z_{n} k_{0} δ_{n}^{4} = 8 π$ , the relative exact-to-paraxial error tends to zero:

$\frac{∥ u_{e x a c t, n} (z_{n}) - u_{p a r, n} (z_{n}) ∥_{2}}{∥ u_{0, n} ∥_{2}} ⟶ 0.$

Statement references

Forward angular-spectrum beam

Paraxial phase and longitudinal-field bounds

$\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$

The complex exponential by its power series

Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive

Facts & Assumptions

Given: The scalar propagators and Fourier convention of the angular-spectrum Definition; countable choice as assumed by its $L^{2}$ Plancherel supplier.

[F1]

The exact and paraxial Fourier multipliers are respectively $e^{i z k_{0} \sqrt{1 - s}}$ and $e^{i z k_{0} (1 - s / 2)}$ , where $s = ∣ 2 π ξ ∣^{2} / k_{0}^{2}$ . Forward angular-spectrum beam

[F2]

The exact remainder is $R (s) = s^{2} / [4 (1 - s / 2 + \sqrt{1 - s})]$ , and for real phases $a, b$ the local lemma gives $∣ e^{i a} - e^{i b} ∣ \leq ∣ a - b ∣$ . Uniform phase-multiplier bound for the square-root dispersion law

[F3]

The Fourier transform $F_{2}$ is unitary on complex $L^{2}$ under countable choice. Plancherel theorem The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F4]

Under countable choice, each nondegenerate box has Lebesgue measure equal to the product of its side lengths. A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included Axis-parallel rectangles in $R^{m}$ and their volume The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F5]

The complex exponential addition law holds; its power-series definition gives $e^{0} = 1$ . The Cartesian/modulus corollary states $∣ e^{i t} ∣ = e^{0}$ for real $t$ , so $∣ e^{i t} ∣ = 1$ ; it also gives Euler's identity $e^{i π} = - 1$ . Consequently, for $Δ \in R$ , $∣ e^{- i Δ} - 1 ∣ = ∣ e^{i Δ} - 1 ∣ .$ Indeed, the addition law gives $e^{- i Δ} e^{i Δ} = e^{0} = 1$ , so $e^{- i Δ} - 1 = e^{- i Δ} (1 - e^{i Δ})$ ; modulus multiplicativity and $∣ e^{- i Δ} ∣ = 1$ give the identity. For any real $b$ , the addition law also gives $e^{i (b - Δ)} - e^{i b} = e^{i b} (e^{- i Δ} - 1)$ , so the modulus of this phase-multiplier difference is $∣ e^{i Δ} - 1 ∣$ . $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive

[F6]

The A-page estimate is proportional to the scaled distance $∣ z ∣ k_{0} δ^{4}$ ; it makes no convergence claim when that quantity is fixed and nonzero. Paraxial phase and longitudinal-field bounds

[F7]

Complex modulus is subadditive and multiplicative. For $u, v \in C$ , $∣ u ∣ \leq ∣ u - v ∣ + ∣ v ∣$ and $∣ v ∣ \leq ∣ v - u ∣ + ∣ u ∣ = ∣ u - v ∣ + ∣ u ∣$ , so $∣ ∣ u ∣ - ∣ v ∣ ∣ \leq ∣ u - v ∣$ ; the equality $∣ v - u ∣ = ∣ u - v ∣$ follows from $v - u = - (u - v)$ , multiplicativity, and $∣ - 1 ∣ = 1$ from the modulus definition. Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive

Counterexample

Given: Fix $k_{0} > 0$ and choose one constant spectral norm $A_{⋆} > 0$ (in volts for scalar fields measured in $V m^{- 1}$ ).

1.1givenF4algebra

Set $δ_{n} : = 1 / (n + 2)$ and $R_{n} : = δ_{n} k_{0} / (2 π)$ . Let $Ξ_{n} : = [R_{n} (1 - δ_{n} / 4), R_{n} (1 - δ_{n} / 8)] \times [0, R_{n} δ_{n} / 8]$ . Its side lengths are both $R_{n} δ_{n} / 8 > 0$ , so [F4] gives $0 < ∣ Ξ_{n} ∣ = R_{n}^{2} δ_{n}^{2} / 64 < \infty$ . Define ${\hat{u}}_{0, n} : = A_{⋆} ∣ Ξ_{n} ∣^{- 1 / 2} 1_{Ξ_{n}}$ ; then $∥ {\hat{u}}_{0, n} ∥_{2} = A_{⋆}$ . For $q = 2 π ξ$ with $ξ \in Ξ_{n}$ , $q_{x} \geq δ_{n} k_{0} (1 - δ_{n} / 4) > δ_{n} k_{0} (1 - δ_{n})$ and $∣ q ∣^{2} \leq δ_{n}^{2} k_{0}^{2} [(1 - δ_{n} / 8)^{2} + (δ_{n} / 8)^{2}] \leq δ_{n}^{2} k_{0}^{2}$ ; hence this nonzero spectrum lies in the angular cone and concentrates near its outer radius.

2.1step 1.1F3F6algebra

Let $u_{0, n} : = F_{2}^{- 1} {\hat{u}}_{0, n}$ and $z_{n} : = 8 π / (k_{0} δ_{n}^{4}) > 0$ . This fixes the scale in [F6] at $8 π$ . By [F3], $∥ u_{0, n} ∥_{2} = A_{⋆}$ , so each input is nonzero and admissible for the refuted statement.

2.2step 1.1F2algebra

For $ξ \in Ξ_{n}$ , put $s_{n} : = ∣ 2 π ξ ∣^{2} / k_{0}^{2}$ . Step 1.1 gives $(1 - δ_{n} / 4)^{2} \leq s_{n} / δ_{n}^{2} \leq 1$ , so this ratio tends to $1$ uniformly on $Ξ_{n}$ . Also $0 \leq s_{n} \leq δ_{n}^{2}$ , and $∣ 1 - \sqrt{1 - s_{n}} ∣ = s_{n} / (1 + \sqrt{1 - s_{n}}) \leq s_{n}$ , so $1 - s_{n} / 2 + \sqrt{1 - s_{n}} \to 2$ uniformly. By [F2], the relative phase difference is $Δ_{n} : = z_{n} k_{0} R (s_{n}) = 2 π (s_{n} / δ_{n}^{2})^{2} / (1 - s_{n} / 2 + \sqrt{1 - s_{n}})$ , which tends uniformly to $π$ .

3.1step 2.2F2F5F7algebra

The actual exact-minus-paraxial phase is $- Δ_{n}$ , so [F5] makes the modulus of the multiplier difference $∣ e^{i Δ_{n}} - 1 ∣$ . Put $u = e^{i Δ_{n}} - 1$ and $v = e^{i π} - 1$ . By [F7], $∣ ∣ u ∣ - ∣ v ∣ ∣ \leq ∣ u - v ∣$ ; [F5] gives $∣ v ∣ = 2$ , and [F2] gives $∣ u - v ∣ = ∣ e^{i Δ_{n}} - e^{i π} ∣ \leq ∣ Δ_{n} - π ∣ \to 0$ uniformly on $Ξ_{n}$ . Thus the multiplier-difference modulus tends uniformly to $2$ .

4.1step 1.1step 2.1step 3.1F1F3algebra∎

By [F3], the squared relative $L^{2}$ error is the average of $∣ e^{i Δ_{n}} - 1 ∣^{2}$ over $Ξ_{n}$ with respect to normalized Lebesgue measure. Since step 3.1 gives uniform convergence of that modulus to $2$ , the averages of its square tend to $4$ ; the relative errors are nonnegative, so they tend to $2$ , not $0$ . This contradicts the Statement refuted and proves the claimed counterexample.

A small angular cone can accumulate order-one phase error

Statement refuted

Statement references

Facts & Assumptions

Counterexample

Depends on

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