Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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Fundamental solution by division of a fourier symbol

Example

Assume Countable Choice and write D=d/dx. The integrable function

E(x)=12ex

defines a tempered fundamental solution for 1D2 on R:

(1D2)E=δ0,FE(ξ)=11+4π2ξ2.

Facts & Assumptions

Given: Countable Choice and the 2π Fourier convention.

[F1]

The distributional transform of an L1 function is represented by its integral transform (Fourier transform agrees with l one and plancherel transforms).

[F2]

The symbol identity is F(P(D)u)=P(2πiξ)Fu (Constant coefficient differential operators become polynomial multipliers).

Verification

technique · explicit integral transform and symbol multiplication
1.1

Since EL1(R), [F1] applies. Split at zero and use the elementary decaying exponential antiderivative.

F1algebra

E^(ξ)=12(0e(1+2πiξ)xdx+0e(12πiξ)xdx)=12(11+2πiξ+112πiξ)=11+4π2ξ2.

[F1, algebra]

2.1

For P(z)=1z2, apply the Fourier-symbol identity to step 1.1.

F2step 1.1

F((1D2)E)=(1(2πiξ)2)E^(ξ)=1.

But 1=Fδ0 by [F3], so injectivity yields (1D2)E=δ0. [F2, F3, step 1.1]

3.1

The denominator is strictly positive on the real frequency axis, so this particular division produces a smooth bounded multiplier. The calculation does not assert that an arbitrary polynomial symbol can be divided in S, nor any general PDE existence or regularity theorem. Countable Choice is used only through [F1]–[F3].

givenstep 2.1

Depends on

Used by

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Sources