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Principal value one over x is tempered and its fourier transform

Example

Assume Countable Choice. The principal-value functional

$⟨ pv \frac{1}{x}, φ ⟩ = \lim_{ε ↓ 0} \int_{∣ x ∣ > ε} \frac{φ (x)}{x} d x$

is a tempered distribution on $R$ , and

$F (pv \frac{1}{x}) (ξ) = - i π sgn (ξ)$

in $S^{'} (R)$ for the negative-sign $2 π$ normalization.

Facts & Assumptions

Given: Countable Choice and $φ \in S (R)$ .

[F1]

A finite Schwartz-seminorm estimate characterizes tempered distributions, while its restriction gives the local finite-order condition on compact tests (Finite seminorm bound characterizes tempered distributions, Local finite order characterization of distributions).

[F2]

Multiplication by $x$ and Fourier multiplication/differentiation have the published distributional meanings and exact constants (Multiplication of a distribution by a smooth function, Fourier differentiation and multiplication identities on tempered distributions).

[F3]

The transforms of $1$ and $δ_{0}$ are known with the $2 π$ normalization (Fourier transform of delta constants plane waves and polynomials).

[F4]

Restriction $S^{'} \to D^{'}$ is injective, and a distribution with zero derivative on connected $R$ is constant (Tempered distributions embed continuously in distributions, A distribution with zero derivatives on a connected open set is constant).

[F5]

Complex integration by parts on decaying lines and the integral triangle inequality are available (Complex integration by parts on intervals and decaying lines, The modulus of an integral is bounded by the integral of the modulus).

Verification

technique · direct principal-value bound and a distributional ODE

1.1

Symmetry cancels the constant term near zero, reducing the defining limit to two absolutely convergent integrals.

F1F5

$\int_{0 < ∣ x ∣ < 1} \frac{φ (x) - φ (0)}{x} d x + \int_{∣ x ∣ \geq 1} \frac{φ (x)}{x} d x .$

Both integrals are absolute. The mean-value estimate and Schwartz decay give

$∣ ⟨ pv \frac{1}{x}, φ ⟩ ∣ \leq 2 p_{0, 1} (φ) + p_{2, 0} (φ),$

because $\int_{∣ x ∣ \geq 1} ∣ x ∣^{- 3} d x = 1$ . Thus the limit exists, [F1] proves temperateness, and the same estimate restricts to a finite-order $D^{'}$ functional. [F1, F5]

2.1

Test multiplication by the smooth coordinate function $x$ .

F2step 1.1

$⟨ x pv \frac{1}{x}, φ ⟩ = \lim_{ε ↓ 0} \int_{∣ x ∣ > ε} φ (x) d x = \int φ (x) d x .$

Hence $x pv (1 / x) = 1$ . [F2, step 1.1]

3.1

Put $U = F (pv (1 / x))$ . Transform step 2.1 and use the exact Fourier multiplication law and constant transform.

F2F3step 2.1

$- \frac{1}{2 π i} U^{'} = δ_{0}, U^{'} = - 2 π i δ_{0} .$

[F2, F3, step 2.1]

4.1

Integration by parts on the two half-lines gives $(sgn)^{'} = 2 δ_{0}$ : indeed $- \int sgn (x) ψ^{'} (x) d x = 2 ψ (0)$ for every compact test $ψ$ . Thus $S = - i π sgn$ satisfies $S^{'} = - 2 π i δ_{0}$ , and $W = U - S$ has zero derivative. The bounded function $sgn$ is itself regular tempered by the elementary estimate $∣ \int sgn φ ∣ \leq \int ∣ φ ∣ \leq C p_{2, 0} (φ)$ .

F1F5step 3.1

5.1

By [F4], the restriction of $W$ is a constant distribution $c$ . The principal value is odd under reflection, Fourier transformation commutes with reflection by its defining integral, and $S$ is odd; hence $W$ is odd. A constant distribution is even, so $c = - c$ and $c = 0$ . Injectivity in [F4] then makes $W = 0$ already in $S^{'}$ . This proves $U = S$ . Countable Choice is used only through the cited Fourier, integration, and zero-derivative interfaces.

F4step 4.1∎

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Sources

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