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The Hilbert transform is an L2 isometry and squares to minus the identity

Statement

Assume Countable Choice, use the $e^{- 2 π i x ξ}$ convention, and let $m (ξ) = - i sgn (ξ)$ with $sgn (0) = 0$ . The Hilbert transform of The Hilbert transform is the tempered convolution with pv(1/(pi x)) and has signum Fourier multiplier defines, on Schwartz functions, the operator $H f = (W * f)$ with $F (H f) = m \hat{f}$ . Then $H$ extends uniquely to a bounded operator on $L^{2} (R; C)$ , still denoted $H$ , and for every $f \in L^{2} (R; C)$ ,

$∥ H f ∥_{2} = ∥ f ∥_{2}, H^{2} f = - f .$

In particular the single point $ξ = 0$ , where $m$ vanishes, is a Lebesgue-null set and creates no zero-mode exception. Nonzero constant functions are not in $L^{2} (R)$ , so there is no constant mode in the domain to transform.

Facts & Assumptions

Given: Countable Choice and the multiplier $m (ξ) = - i sgn (ξ)$ of the Schwartz Hilbert transform.

[F1]

For Schwartz $f$ the principal-value Hilbert transform $H f = W * f$ satisfies $F (H f) = - i sgn (ξ) \hat{f} (ξ)$ as tempered distributions. The Hilbert transform is the tempered convolution with pv(1/(pi x)) and has signum Fourier multiplier

[F2]

A measurable multiplier $m$ with finite essential supremum defines the bounded operator $T_{m} = F_{2}^{- 1} M_{m} F_{2}$ on $L^{2}$ ; its Schwartz-core action extends uniquely to $L^{2}$ , it depends only on the almost-everywhere class of $m$ , and $∥ T_{m} ∥ = ∥ m ∥_{\infty}$ . Exact L2 Fourier multiplier norm

[F3]

Plancherel: $F_{2}$ is a surjective linear isometry of $L^{2}$ , so $∥ F_{2} g ∥_{2} = ∥ g ∥_{2}$ and $F_{2}^{- 1} (- g) = - F_{2}^{- 1} g$ and $F_{2}^{- 1} (- F_{2} g) = - g$ . Plancherel theorem

[F4]

Fourier transformation is injective on tempered distributions. Fourier transform is a topological automorphism of tempered distributions

Proof

technique · direct

1.1givenalgebra

The symbol satisfies $∣ m (ξ) ∣ = 1$ for every $ξ \neq 0$ , $m (ξ)^{2} = - 1$ for every $ξ \neq 0$ , and the exceptional set is the Lebesgue-null singleton ${0}$ .

2.1step 1.1F1F2F4

For Schwartz $f$ , [F2] identifies $T_{m} f$ with an $L^{2}$ class whose regular tempered distribution has Fourier transform $m \hat{f}$ . By [F1], the tempered distribution $H f$ has the same transform. Injectivity [F4] gives equality of these distributions, so $H f$ is represented by the $L^{2}$ class $T_{m} f$ . Thus no $L^{2}$ membership of the principal value is assumed in this identification.

3.1step 1.1step 2.1F2F3

By [F2] the Schwartz-core action of $T_{m}$ extends uniquely to a bounded operator on $L^{2}$ ; by 2.1 the Schwartz action of $H$ is that core action, so $H = T_{m}$ on $L^{2}$ , and for $g \in L^{2}$ , $∥ H g ∥_{2} = ∥ m F_{2} g ∥_{2} = ∥ F_{2} g ∥_{2} = ∥ g ∥_{2}$ because $∣ m ∣ = 1$ almost everywhere by 1.1.

4.1step 1.1step 2.1F2F3∎

Likewise, on the Schwartz core $H^{2} f = T_{m}^{2} f = F_{2}^{- 1} (m^{2} F_{2} f) = F_{2}^{- 1} (- F_{2} f) = - f$ by 1.1 and [F3]; both $H^{2}$ and $- I$ are bounded on $L^{2}$ and agree on the dense Schwartz core, so $H^{2} = - I$ on all of $L^{2}$ .

Depends on

Used by

Hilbert transform does not map L-infinity to L-infinity Counterexample
Hilbert transform is not strong type (1,1) Counterexample
Finite sum of Riesz squares in L2 Example
Hilbert transform of an interval indicator Example
Hilbert transform of the line Poisson kernel Example
The Hilbert transform is skew-adjoint on L2 Lemma
Endpoint map for Hilbert and Riesz transforms Remark

Dependency tree · two levels

38 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Loukas Grafakos, Classical Fourier Analysis, third edition (standard reference, not scraped)
Richard S. Laugesen, Harmonic Analysis Lecture Notes (standard reference, not scraped)