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Fourier transform agrees with l one and plancherel transforms

Statement

Assume Countable Choice and use the negative-sign $2 π$ normalization. If $f \in L^{1} (R^{n}; C)$ , then

$F u_{f} = u_{\hat{f}},$

where $\hat{f}$ is the integral Fourier transform. If $f \in L^{2} (R^{n}; C)$ , then

$F u_{f} = u_{F_{2} f},$

where $F_{2}$ is the Plancherel extension. These equalities are in $S^{'} (R^{n})$ and therefore depend only on the corresponding almost-everywhere classes.

Facts & Assumptions

Given: Countable Choice and the fixed $2 π$ Fourier convention.

[F1]

Every $L^{p}$ class, including $p = 1, 2, \infty$ , defines a regular tempered distribution (Polynomial growth functions define tempered distributions).

[F2]

The transform on $S^{'}$ is defined by bilinear transposition (Fourier transform of a tempered distribution).

[F3]

Absolute Fubini applies on the sigma-finite Euclidean product (Fubini's theorem for L^1 functions on a sigma-finite product).

[F4]

Schwartz space is dense in complex $L^{2}$ , and the Plancherel transform is its unitary extension (Schwartz space is dense in L2, Plancherel theorem).

[F5]

The integral and Plancherel transforms agree on $L^{1} \cap L^{2}$ (Agreement of the integral and L2 transforms).

[F6]

Hölder applied to moduli controls all $L^{2}$ test pairings (Holder's inequality for integrals, including the endpoint cases).

Proof

technique · Fubini followed by $L^2$ approximation

1.1

Let $f \in L^{1}$ and $φ \in S$ . Schwartz decay makes $φ \in L^{1}$ , so $\int ⁣ \int ∣ f (x) φ (ξ) ∣ d x d ξ < \infty$ . Absolute Fubini is therefore applicable.

F2F3

$⟨ F u_{f}, φ ⟩ = \int f (x) ⁣ (\int φ (ξ) e^{- 2 π i x \cdot ξ} d ξ) d x = \int \hat{f} (ξ) φ (ξ) d ξ .$

Since $\hat{f}$ is bounded, [F1] makes the last functional tempered. This proves the $L^{1}$ assertion. [F1, F2, F3]

1.2

Let $f \in L^{2}$ and choose $f_{j} \in S$ with $f_{j} \to f$ in $L^{2}$ . For a fixed $φ \in S$ , also $F φ \in L^{2}$ , and Hölder gives the first convergence below.

F4F6

$\int (f_{j} - f) F φ ⟶ 0.$

Plancherel gives $F_{2} f_{j} \to F_{2} f$ in $L^{2}$ , so a second Hölder estimate gives $\int (F_{2} f_{j} - F_{2} f) φ \to 0$ . [F4, F6]

2.1

Each $f_{j}$ belongs to $L^{1} \cap L^{2}$ , and the two agreement results give the displayed identity.

F5step 1.1

$\int f_{j} F φ = \int (F_{2} f_{j}) φ .$

Passing to the two limits from step 1.2 yields $⟨ F u_{f}, φ ⟩ = ⟨ u_{F_{2} f}, φ ⟩$ . Since this holds for every Schwartz test, the $L^{2}$ assertion follows. [F1, F2, F5, step 1.2] ∎

Depends on

Used by

Fundamental solution by division of a fourier symbol Example

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Sources

Semyon Dyatlov, Lecture notes for 18.155 (2022) (standard reference, not scraped)