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Riesz transforms are L2 contractions and square to minus the identity in sum

Statement

Assume Countable Choice and let $n \geq 1$ . For the Riesz transforms $R_{1}, \dots, R_{n}$ of the multiplier definition,

$∥ R_{j} f ∥_{2} \leq ∥ f ∥_{2} (j = 1, \dots, n, f \in L^{2} (R^{n}; C)),$

and

$\sum_{j = 1}^{n} R_{j}^{2} f = - f (f \in L^{2} (R^{n}; C)) .$

Both statements are $L^{2}$ statements only; no $L^{p}$ bound for $p \neq 2$ is asserted, and the operators are the $L^{2}$ operators of the definition, so all identities hold as classes (no pointwise statement is made).

Facts & Assumptions

Given: Countable Choice, the dimension $n \geq 1$ , and the Riesz transforms $R_{j} = F_{2}^{- 1} M_{m_{j}} F_{2}$ with symbols $m_{j} (ξ) = - i ξ_{j} / ∣ ξ ∣$ for $ξ \neq 0$ and $m_{j} (0) = 0$ .

[F1]

Each $R_{j}$ is defined as the bounded $L^{2}$ operator with multiplier $m_{j}$ ; the symbol is measurable with $∣ m_{j} (ξ) ∣ \leq 1$ everywhere, the value at the origin is immaterial, and the definition asserts no more than the multiplier description. Riesz transforms on Euclidean space

[F2]

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

[F3]

Plancherel: $F_{2}$ is a surjective complex-linear isometry of $L^{2} (R^{n}; C)$ , so $∥ F_{2} g ∥_{2} = ∥ g ∥_{2}$ . Its inverse is complex-linear, hence $F_{2}^{- 1} (- g) = - F_{2}^{- 1} g$ and $F_{2}^{- 1} (- F_{2} g) = - g$ for every class $g$ . Plancherel theorem

Proof

technique · direct

1.1F1F2

For every $ξ \neq 0$ the symbol values satisfy $\sum_{j = 1}^{n} m_{j} (ξ)^{2} = \sum_{j = 1}^{n} (- ξ_{j}^{2} / ∣ ξ ∣^{2}) = - 1$ , while $m_{j} (0) = 0$ ; the single point ${0}$ is Lebesgue null. Hence the function $s (ξ) : = \sum_{j = 1}^{n} m_{j} (ξ)^{2}$ is measurable, bounded with $∣ s ∣ \leq 1$ , and equals the constant $- 1$ almost everywhere.

1.2F1F2F3

By [F2] applied to the bounded measurable symbol $m_{j}$ of [F1], $∥ R_{j} ∥ = ∥ m_{j} ∥_{\infty} \leq 1$ ; consequently, for $f \in L^{2}$ and using the isometry of [F3], $∥ R_{j} f ∥_{2} = ∥ m_{j} F_{2} f ∥_{2} \leq ∥ F_{2} f ∥_{2} = ∥ f ∥_{2}$ .

2.1step 1.1F2F3

Since $R_{j} = F_{2}^{- 1} M_{m_{j}} F_{2}$ , composition gives $R_{j}^{2} = F_{2}^{- 1} M_{m_{j}^{2}} F_{2} = T_{m_{j}^{2}}$ in the notation of [F2], and summing the finitely many bounded operators gives $\sum_{j = 1}^{n} R_{j}^{2} = T_{s}$ for the almost-everywhere- $- 1$ symbol $s$ of 1.1.

3.1step 1.1step 2.1F2F3∎

By [F2] the operator $T_{s}$ depends only on the almost-everywhere class of $s$ , which by 1.1 is the class of the constant $- 1$ ; hence $T_{s} = T_{- 1}$ , and $T_{- 1} f = F_{2}^{- 1} M_{- 1} F_{2} f = F_{2}^{- 1} (- F_{2} f) = - f$ by the linearity and isometry of [F3]. Therefore $\sum_{j = 1}^{n} R_{j}^{2} f = - f$ for every $f \in L^{2} (R^{n}; C)$ .

Depends on

Used by

Finite sum of Riesz squares in L2 Example
Endpoint map for Hilbert and Riesz transforms Remark

Dependency tree · two levels

27 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)